Seeking Methods to solve $Fleft(alpharight) = int_{0}^{1} x^alpha arcsin(x):dx$












5












$begingroup$


I'm looking for different methods to solve the following integral.
$$ Fleft(alpharight) = int_{0}^{1} x^alpha arcsin(x):dx$$



For $alpha > 0$



Here the method I took was to employ integration by parts and then call to special functions, but can this equally be achieved with say a Feynman Trick? or another form integral transform?



My approach in detail:



Employ integration by parts:



begin{align}
v'(x) &= x^alpha & u(x) &= arcsin(x) \
v(x) &= frac{x^{alpha + 1}}{alpha + 1} & u'(x) &= frac{1}{sqrt{1 - x^2}}
end{align}



Thus,



begin{align}
Fleft(alpharight) &= left[frac{x^{alpha + 1}}{alpha + 1}cdotarcsin(x)right]_0^1 - int_0^1 frac{x^{alpha + 1}}{alpha + 1} cdot frac{1}{sqrt{1 - x^2}} :dx \
&= frac{pi}{2left(alpha + 1right)} - frac{1}{alpha + 1}int_0^1 x^{alpha + 1}left(1 - x^2right)^{-frac{1}{2}} :dx
end{align}



Here make the substitution $u = x^2$ to obtain



begin{align}
Fleft(alpharight) &= frac{pi}{2left(alpha + 1right)} - frac{1}{alpha + 1}int_0^1 left(sqrt{u}right)^{alpha + 1}left(1 - uright)^{-frac{1}{2}} frac{:du}{2sqrt{u}} \
&= frac{pi}{2left(alpha + 1right)} - frac{1}{2left(alpha + 1right)}int_0^1 u^{frac{alpha}{2}}left(1 - uright) ^{-frac{1}{2}} :du \
&= frac{1}{2left(alpha + 1right)} left[ pi - Bleft(frac{alpha + 2}{2}, frac{1}{2} right) right]
end{align}



begin{align}
Fleft(alpharight) &=frac{1}{2left(alpha + 1right)} left[ pi - frac{Gammaleft(frac{alpha + 2}{2}right)Gammaleft(frac{1}{2}right)}{Gammaleft(frac{alpha + 2}{2} + frac{1}{2}right)} right] \
&= frac{1}{2left(alpha + 1right)} left[ pi - frac{Gammaleft(frac{alpha + 2}{2}right)sqrt{pi}}{Gammaleft(frac{alpha + 3}{2}right) } right] \
&= frac{sqrt{pi}}{2left(alpha + 1right)} left[ sqrt{pi} - frac{Gammaleft(frac{alpha + 2}{2}right)}{Gammaleft(frac{alpha + 3}{2}right) } right]
end{align}



Edits:
Correction of original limit observation (now removed)
Correction of not stating region of convergence for $alpha$.
Correction of 1/sqrt to sqrt in final line.



Thanks to those commentators for pointing out.






integration definite-integrals error-function beta-function gamma-distribution







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How would you expect to get $Gamma$'s out of Feynman Trick or whatever else? (And, BTW, $displaystylelim_{alphatoinfty}F(alpha)=0$).
    $endgroup$
    – metamorphy
    Dec 10 '18 at 1:03






  • 2




    $begingroup$
    There needs to be some restrictions on the value of $alpha$ otherwise the improper integral will not converge, namely $alpha > -2$. The case $alpha = -1$ also needs to be handled with care.
    $endgroup$
    – omegadot
    Dec 10 '18 at 2:12






  • 1




    $begingroup$
    Yes, sorry. I meant to say it only converges for $alpha > -2$ and diverges for $alpha leqslant -2$.
    $endgroup$
    – omegadot
    Dec 10 '18 at 3:02






  • 1




    $begingroup$
    I'm fairly certain that's the fastest way to do it
    $endgroup$
    – clathratus
    Dec 10 '18 at 3:29






  • 1




    $begingroup$
    The $frac{1}{sqrt{pi }}$ in the last line needs to be $sqrt{pi}$.
    $endgroup$
    – JimB
    Dec 10 '18 at 5:12
















5












$begingroup$


I'm looking for different methods to solve the following integral.
$$ Fleft(alpharight) = int_{0}^{1} x^alpha arcsin(x):dx$$



For $alpha > 0$



Here the method I took was to employ integration by parts and then call to special functions, but can this equally be achieved with say a Feynman Trick? or another form integral transform?



My approach in detail:



Employ integration by parts:



begin{align}
v'(x) &= x^alpha & u(x) &= arcsin(x) \
v(x) &= frac{x^{alpha + 1}}{alpha + 1} & u'(x) &= frac{1}{sqrt{1 - x^2}}
end{align}



Thus,



begin{align}
Fleft(alpharight) &= left[frac{x^{alpha + 1}}{alpha + 1}cdotarcsin(x)right]_0^1 - int_0^1 frac{x^{alpha + 1}}{alpha + 1} cdot frac{1}{sqrt{1 - x^2}} :dx \
&= frac{pi}{2left(alpha + 1right)} - frac{1}{alpha + 1}int_0^1 x^{alpha + 1}left(1 - x^2right)^{-frac{1}{2}} :dx
end{align}



Here make the substitution $u = x^2$ to obtain



begin{align}
Fleft(alpharight) &= frac{pi}{2left(alpha + 1right)} - frac{1}{alpha + 1}int_0^1 left(sqrt{u}right)^{alpha + 1}left(1 - uright)^{-frac{1}{2}} frac{:du}{2sqrt{u}} \
&= frac{pi}{2left(alpha + 1right)} - frac{1}{2left(alpha + 1right)}int_0^1 u^{frac{alpha}{2}}left(1 - uright) ^{-frac{1}{2}} :du \
&= frac{1}{2left(alpha + 1right)} left[ pi - Bleft(frac{alpha + 2}{2}, frac{1}{2} right) right]
end{align}



begin{align}
Fleft(alpharight) &=frac{1}{2left(alpha + 1right)} left[ pi - frac{Gammaleft(frac{alpha + 2}{2}right)Gammaleft(frac{1}{2}right)}{Gammaleft(frac{alpha + 2}{2} + frac{1}{2}right)} right] \
&= frac{1}{2left(alpha + 1right)} left[ pi - frac{Gammaleft(frac{alpha + 2}{2}right)sqrt{pi}}{Gammaleft(frac{alpha + 3}{2}right) } right] \
&= frac{sqrt{pi}}{2left(alpha + 1right)} left[ sqrt{pi} - frac{Gammaleft(frac{alpha + 2}{2}right)}{Gammaleft(frac{alpha + 3}{2}right) } right]
end{align}



Edits:
Correction of original limit observation (now removed)
Correction of not stating region of convergence for $alpha$.
Correction of 1/sqrt to sqrt in final line.



Thanks to those commentators for pointing out.






integration definite-integrals error-function beta-function gamma-distribution







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How would you expect to get $Gamma$'s out of Feynman Trick or whatever else? (And, BTW, $displaystylelim_{alphatoinfty}F(alpha)=0$).
    $endgroup$
    – metamorphy
    Dec 10 '18 at 1:03






  • 2




    $begingroup$
    There needs to be some restrictions on the value of $alpha$ otherwise the improper integral will not converge, namely $alpha > -2$. The case $alpha = -1$ also needs to be handled with care.
    $endgroup$
    – omegadot
    Dec 10 '18 at 2:12






  • 1




    $begingroup$
    Yes, sorry. I meant to say it only converges for $alpha > -2$ and diverges for $alpha leqslant -2$.
    $endgroup$
    – omegadot
    Dec 10 '18 at 3:02






  • 1




    $begingroup$
    I'm fairly certain that's the fastest way to do it
    $endgroup$
    – clathratus
    Dec 10 '18 at 3:29






  • 1




    $begingroup$
    The $frac{1}{sqrt{pi }}$ in the last line needs to be $sqrt{pi}$.
    $endgroup$
    – JimB
    Dec 10 '18 at 5:12














5












5








5


1



$begingroup$


I'm looking for different methods to solve the following integral.
$$ Fleft(alpharight) = int_{0}^{1} x^alpha arcsin(x):dx$$



For $alpha > 0$



Here the method I took was to employ integration by parts and then call to special functions, but can this equally be achieved with say a Feynman Trick? or another form integral transform?



My approach in detail:



Employ integration by parts:



begin{align}
v'(x) &= x^alpha & u(x) &= arcsin(x) \
v(x) &= frac{x^{alpha + 1}}{alpha + 1} & u'(x) &= frac{1}{sqrt{1 - x^2}}
end{align}



Thus,



begin{align}
Fleft(alpharight) &= left[frac{x^{alpha + 1}}{alpha + 1}cdotarcsin(x)right]_0^1 - int_0^1 frac{x^{alpha + 1}}{alpha + 1} cdot frac{1}{sqrt{1 - x^2}} :dx \
&= frac{pi}{2left(alpha + 1right)} - frac{1}{alpha + 1}int_0^1 x^{alpha + 1}left(1 - x^2right)^{-frac{1}{2}} :dx
end{align}



Here make the substitution $u = x^2$ to obtain



begin{align}
Fleft(alpharight) &= frac{pi}{2left(alpha + 1right)} - frac{1}{alpha + 1}int_0^1 left(sqrt{u}right)^{alpha + 1}left(1 - uright)^{-frac{1}{2}} frac{:du}{2sqrt{u}} \
&= frac{pi}{2left(alpha + 1right)} - frac{1}{2left(alpha + 1right)}int_0^1 u^{frac{alpha}{2}}left(1 - uright) ^{-frac{1}{2}} :du \
&= frac{1}{2left(alpha + 1right)} left[ pi - Bleft(frac{alpha + 2}{2}, frac{1}{2} right) right]
end{align}



begin{align}
Fleft(alpharight) &=frac{1}{2left(alpha + 1right)} left[ pi - frac{Gammaleft(frac{alpha + 2}{2}right)Gammaleft(frac{1}{2}right)}{Gammaleft(frac{alpha + 2}{2} + frac{1}{2}right)} right] \
&= frac{1}{2left(alpha + 1right)} left[ pi - frac{Gammaleft(frac{alpha + 2}{2}right)sqrt{pi}}{Gammaleft(frac{alpha + 3}{2}right) } right] \
&= frac{sqrt{pi}}{2left(alpha + 1right)} left[ sqrt{pi} - frac{Gammaleft(frac{alpha + 2}{2}right)}{Gammaleft(frac{alpha + 3}{2}right) } right]
end{align}



Edits:
Correction of original limit observation (now removed)
Correction of not stating region of convergence for $alpha$.
Correction of 1/sqrt to sqrt in final line.



Thanks to those commentators for pointing out.






integration definite-integrals error-function beta-function gamma-distribution







share|cite|improve this question











$endgroup$




I'm looking for different methods to solve the following integral.
$$ Fleft(alpharight) = int_{0}^{1} x^alpha arcsin(x):dx$$



For $alpha > 0$



Here the method I took was to employ integration by parts and then call to special functions, but can this equally be achieved with say a Feynman Trick? or another form integral transform?



My approach in detail:



Employ integration by parts:



begin{align}
v'(x) &= x^alpha & u(x) &= arcsin(x) \
v(x) &= frac{x^{alpha + 1}}{alpha + 1} & u'(x) &= frac{1}{sqrt{1 - x^2}}
end{align}



Thus,



begin{align}
Fleft(alpharight) &= left[frac{x^{alpha + 1}}{alpha + 1}cdotarcsin(x)right]_0^1 - int_0^1 frac{x^{alpha + 1}}{alpha + 1} cdot frac{1}{sqrt{1 - x^2}} :dx \
&= frac{pi}{2left(alpha + 1right)} - frac{1}{alpha + 1}int_0^1 x^{alpha + 1}left(1 - x^2right)^{-frac{1}{2}} :dx
end{align}



Here make the substitution $u = x^2$ to obtain



begin{align}
Fleft(alpharight) &= frac{pi}{2left(alpha + 1right)} - frac{1}{alpha + 1}int_0^1 left(sqrt{u}right)^{alpha + 1}left(1 - uright)^{-frac{1}{2}} frac{:du}{2sqrt{u}} \
&= frac{pi}{2left(alpha + 1right)} - frac{1}{2left(alpha + 1right)}int_0^1 u^{frac{alpha}{2}}left(1 - uright) ^{-frac{1}{2}} :du \
&= frac{1}{2left(alpha + 1right)} left[ pi - Bleft(frac{alpha + 2}{2}, frac{1}{2} right) right]
end{align}



begin{align}
Fleft(alpharight) &=frac{1}{2left(alpha + 1right)} left[ pi - frac{Gammaleft(frac{alpha + 2}{2}right)Gammaleft(frac{1}{2}right)}{Gammaleft(frac{alpha + 2}{2} + frac{1}{2}right)} right] \
&= frac{1}{2left(alpha + 1right)} left[ pi - frac{Gammaleft(frac{alpha + 2}{2}right)sqrt{pi}}{Gammaleft(frac{alpha + 3}{2}right) } right] \
&= frac{sqrt{pi}}{2left(alpha + 1right)} left[ sqrt{pi} - frac{Gammaleft(frac{alpha + 2}{2}right)}{Gammaleft(frac{alpha + 3}{2}right) } right]
end{align}



Edits:
Correction of original limit observation (now removed)
Correction of not stating region of convergence for $alpha$.
Correction of 1/sqrt to sqrt in final line.



Thanks to those commentators for pointing out.






integration definite-integrals error-function beta-function gamma-distribution




integration definite-integrals error-function beta-function gamma-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 10 '18 at 5:14







DavidG

















asked Dec 10 '18 at 0:37









DavidGDavidG

1




1








  • 1




    $begingroup$
    How would you expect to get $Gamma$'s out of Feynman Trick or whatever else? (And, BTW, $displaystylelim_{alphatoinfty}F(alpha)=0$).
    $endgroup$
    – metamorphy
    Dec 10 '18 at 1:03






  • 2




    $begingroup$
    There needs to be some restrictions on the value of $alpha$ otherwise the improper integral will not converge, namely $alpha > -2$. The case $alpha = -1$ also needs to be handled with care.
    $endgroup$
    – omegadot
    Dec 10 '18 at 2:12






  • 1




    $begingroup$
    Yes, sorry. I meant to say it only converges for $alpha > -2$ and diverges for $alpha leqslant -2$.
    $endgroup$
    – omegadot
    Dec 10 '18 at 3:02






  • 1




    $begingroup$
    I'm fairly certain that's the fastest way to do it
    $endgroup$
    – clathratus
    Dec 10 '18 at 3:29






  • 1




    $begingroup$
    The $frac{1}{sqrt{pi }}$ in the last line needs to be $sqrt{pi}$.
    $endgroup$
    – JimB
    Dec 10 '18 at 5:12














  • 1




    $begingroup$
    How would you expect to get $Gamma$'s out of Feynman Trick or whatever else? (And, BTW, $displaystylelim_{alphatoinfty}F(alpha)=0$).
    $endgroup$
    – metamorphy
    Dec 10 '18 at 1:03






  • 2




    $begingroup$
    There needs to be some restrictions on the value of $alpha$ otherwise the improper integral will not converge, namely $alpha > -2$. The case $alpha = -1$ also needs to be handled with care.
    $endgroup$
    – omegadot
    Dec 10 '18 at 2:12






  • 1




    $begingroup$
    Yes, sorry. I meant to say it only converges for $alpha > -2$ and diverges for $alpha leqslant -2$.
    $endgroup$
    – omegadot
    Dec 10 '18 at 3:02






  • 1




    $begingroup$
    I'm fairly certain that's the fastest way to do it
    $endgroup$
    – clathratus
    Dec 10 '18 at 3:29






  • 1




    $begingroup$
    The $frac{1}{sqrt{pi }}$ in the last line needs to be $sqrt{pi}$.
    $endgroup$
    – JimB
    Dec 10 '18 at 5:12








1




1




$begingroup$
How would you expect to get $Gamma$'s out of Feynman Trick or whatever else? (And, BTW, $displaystylelim_{alphatoinfty}F(alpha)=0$).
$endgroup$
– metamorphy
Dec 10 '18 at 1:03




$begingroup$
How would you expect to get $Gamma$'s out of Feynman Trick or whatever else? (And, BTW, $displaystylelim_{alphatoinfty}F(alpha)=0$).
$endgroup$
– metamorphy
Dec 10 '18 at 1:03




2




2




$begingroup$
There needs to be some restrictions on the value of $alpha$ otherwise the improper integral will not converge, namely $alpha > -2$. The case $alpha = -1$ also needs to be handled with care.
$endgroup$
– omegadot
Dec 10 '18 at 2:12




$begingroup$
There needs to be some restrictions on the value of $alpha$ otherwise the improper integral will not converge, namely $alpha > -2$. The case $alpha = -1$ also needs to be handled with care.
$endgroup$
– omegadot
Dec 10 '18 at 2:12




1




1




$begingroup$
Yes, sorry. I meant to say it only converges for $alpha > -2$ and diverges for $alpha leqslant -2$.
$endgroup$
– omegadot
Dec 10 '18 at 3:02




$begingroup$
Yes, sorry. I meant to say it only converges for $alpha > -2$ and diverges for $alpha leqslant -2$.
$endgroup$
– omegadot
Dec 10 '18 at 3:02




1




1




$begingroup$
I'm fairly certain that's the fastest way to do it
$endgroup$
– clathratus
Dec 10 '18 at 3:29




$begingroup$
I'm fairly certain that's the fastest way to do it
$endgroup$
– clathratus
Dec 10 '18 at 3:29




1




1




$begingroup$
The $frac{1}{sqrt{pi }}$ in the last line needs to be $sqrt{pi}$.
$endgroup$
– JimB
Dec 10 '18 at 5:12




$begingroup$
The $frac{1}{sqrt{pi }}$ in the last line needs to be $sqrt{pi}$.
$endgroup$
– JimB
Dec 10 '18 at 5:12










2 Answers
2






active

oldest

votes


















2












$begingroup$

Here is a method that relies on using a double integral.



Noting the integral converges for $alpha > -2$, recognising
$$arcsin x = int_0^x frac{du}{sqrt{1 - u^2}},$$
the integral can be rewritten as
$$I = int_0^1 int_0^x frac{x^alpha}{sqrt{1 - u^2}} , du dx.$$
On changing the order of integration, one has
begin{equation}
int_0^1 int_u^1 frac{x^alpha}{sqrt{1 - u^2}} , dx du. qquad (*)
end{equation}
After performing the $x$-integral we are left with
$$I = frac{1}{alpha + 1} int_0^1 frac{1 - u^{alpha + 1}}{sqrt{1 - u^2}} , du, quad alpha neq -1.$$
Enforcing a substitution of $u mapsto sqrt{u}$ results in
$$I = frac{1}{2(alpha + 1)} int_0^1 left (frac{1}{sqrt{u(1 - u)}} - frac{u^{alpha/2}}{sqrt{ 1 - u}} right ) , du = I_1 - I_2.$$



The first of the integrals is trivial
$$I_1 = frac{1}{2(alpha + 1)} int_0^1 frac{du}{sqrt{frac{1}{4} - (u - frac{1}{2})^2}} = frac{1}{2(alpha + 1)} arcsin (2u - 1) Big{|}^1_0 = frac{pi}{2(alpha + 1)}.$$



For the second of the integrals, it can be evaluated by writing it in terms of the beta function. Here
begin{align}
I_2 &= int_0^1 frac{u^{alpha/2}}{sqrt{1 - u}} , du\
&= int_0^1 u^{(alpha/2 + 1) - 1} (1 - u)^{1/2 - 1} , du\
&= text{B} left (frac{alpha}{2} + 1, frac{1}{2} right )\
&= sqrt{pi} , frac{Gamma left (frac{alpha + 2}{2} right )}{Gamma left (frac{alpha + 3}{2} right )}.\
end{align}
Thus
$$I = frac{1}{2(alpha + 1)} left [pi - sqrt{pi} , frac{Gamma left (frac{alpha + 2}{2} right )}{Gamma left (frac{alpha + 3}{2} right )} right ], qquad alpha neq -1.$$



For the case when $alpha = -1$, the double integral at ($*$) becomes
$$I = int_0^1 int_u^1 frac{1}{xsqrt{1 - u^2}} , dx du.$$
After performing the $x$-integral which yields a natural logarithm, one has
$$I = -int_0^1 frac{ln u}{sqrt{1 - u^2}} , du.$$
Enforcing a substitution of $u mapsto sin u$ leads to
$$I = -int_0^{pi/2} ln (sin u) , du. qquad (**)$$
The integral appearing in ($**$) is quite famous and reasonably (?) well known. Its evaluation can be found either here or here. Thus
$$I = frac{pi}{2} ln 2, qquad alpha = -1.$$






share|cite|improve this answer









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  • $begingroup$
    A fantastic solution. Thank you very much very. I'm been working on developing my general nasty integral skills and this mode of converting a single to double integral is great. Is this a form of Feynman's Trick? Regardless, again, fantastic solution.
    $endgroup$
    – DavidG
    Dec 10 '18 at 12:10



















1












$begingroup$

Answer 2.0:



We know that for $|x|<1$,
$$arcsin x=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)}x^{2k+1}$$
Where $$(1/2)_k=frac{Gamma(1/2+k)}{Gamma(1/2)}$$
Hence we may begin with
$$F(a)=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)}int_0^1x^{2k+a+1}mathrm dx=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)(2k+2+a)}$$
Then we play with the fractions a little to get
$$F(a)=frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}bigg[frac1{2k+1}-frac1{2k+2+a}bigg]$$
$$F(a)=frac1{a+1}sum_{ngeq0}frac{(1/2)_n}{n!}frac1{2n+1}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}frac1{2k+2+a}$$
$$F(a)=fracpi{2(a+1)}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!(2k+2+a)}$$
$$F(a)=fracpi{2(a+1)}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}int_0^1x^{2k+1+a}mathrm dx$$
$$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1sum_{kgeq0}frac{(1/2)_k}{k!}x^{2k+1+a}mathrm dx$$
$$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1frac{x^{a+1}}{Gamma(1/2)}sum_{kgeq0}frac{Gamma(1/2+k)}{k!}x^{2k}mathrm dx$$
$$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1frac{x^{a+1}}{sqrt{1-x^2}}mathrm dx$$
$u=x^2$:
$$F(a)=fracpi{2(a+1)}-frac1{2(a+1)}int_0^1u^{a/2}(1-u)^{-1/2}mathrm du$$
$$F(a)=fracpi{2(a+1)}-frac{Gamma(frac{a+2}2)Gamma(frac12)}{2(a+1)Gamma(frac{a+3}2)}$$
$$F(a)=frac{sqrt{pi}}{2(a+1)}bigg[sqrt{pi}-frac{Gamma(frac{a+2}2)}{Gamma(frac{a+3}2)}bigg]$$






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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Here is a method that relies on using a double integral.



    Noting the integral converges for $alpha > -2$, recognising
    $$arcsin x = int_0^x frac{du}{sqrt{1 - u^2}},$$
    the integral can be rewritten as
    $$I = int_0^1 int_0^x frac{x^alpha}{sqrt{1 - u^2}} , du dx.$$
    On changing the order of integration, one has
    begin{equation}
    int_0^1 int_u^1 frac{x^alpha}{sqrt{1 - u^2}} , dx du. qquad (*)
    end{equation}
    After performing the $x$-integral we are left with
    $$I = frac{1}{alpha + 1} int_0^1 frac{1 - u^{alpha + 1}}{sqrt{1 - u^2}} , du, quad alpha neq -1.$$
    Enforcing a substitution of $u mapsto sqrt{u}$ results in
    $$I = frac{1}{2(alpha + 1)} int_0^1 left (frac{1}{sqrt{u(1 - u)}} - frac{u^{alpha/2}}{sqrt{ 1 - u}} right ) , du = I_1 - I_2.$$



    The first of the integrals is trivial
    $$I_1 = frac{1}{2(alpha + 1)} int_0^1 frac{du}{sqrt{frac{1}{4} - (u - frac{1}{2})^2}} = frac{1}{2(alpha + 1)} arcsin (2u - 1) Big{|}^1_0 = frac{pi}{2(alpha + 1)}.$$



    For the second of the integrals, it can be evaluated by writing it in terms of the beta function. Here
    begin{align}
    I_2 &= int_0^1 frac{u^{alpha/2}}{sqrt{1 - u}} , du\
    &= int_0^1 u^{(alpha/2 + 1) - 1} (1 - u)^{1/2 - 1} , du\
    &= text{B} left (frac{alpha}{2} + 1, frac{1}{2} right )\
    &= sqrt{pi} , frac{Gamma left (frac{alpha + 2}{2} right )}{Gamma left (frac{alpha + 3}{2} right )}.\
    end{align}
    Thus
    $$I = frac{1}{2(alpha + 1)} left [pi - sqrt{pi} , frac{Gamma left (frac{alpha + 2}{2} right )}{Gamma left (frac{alpha + 3}{2} right )} right ], qquad alpha neq -1.$$



    For the case when $alpha = -1$, the double integral at ($*$) becomes
    $$I = int_0^1 int_u^1 frac{1}{xsqrt{1 - u^2}} , dx du.$$
    After performing the $x$-integral which yields a natural logarithm, one has
    $$I = -int_0^1 frac{ln u}{sqrt{1 - u^2}} , du.$$
    Enforcing a substitution of $u mapsto sin u$ leads to
    $$I = -int_0^{pi/2} ln (sin u) , du. qquad (**)$$
    The integral appearing in ($**$) is quite famous and reasonably (?) well known. Its evaluation can be found either here or here. Thus
    $$I = frac{pi}{2} ln 2, qquad alpha = -1.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      A fantastic solution. Thank you very much very. I'm been working on developing my general nasty integral skills and this mode of converting a single to double integral is great. Is this a form of Feynman's Trick? Regardless, again, fantastic solution.
      $endgroup$
      – DavidG
      Dec 10 '18 at 12:10
















    2












    $begingroup$

    Here is a method that relies on using a double integral.



    Noting the integral converges for $alpha > -2$, recognising
    $$arcsin x = int_0^x frac{du}{sqrt{1 - u^2}},$$
    the integral can be rewritten as
    $$I = int_0^1 int_0^x frac{x^alpha}{sqrt{1 - u^2}} , du dx.$$
    On changing the order of integration, one has
    begin{equation}
    int_0^1 int_u^1 frac{x^alpha}{sqrt{1 - u^2}} , dx du. qquad (*)
    end{equation}
    After performing the $x$-integral we are left with
    $$I = frac{1}{alpha + 1} int_0^1 frac{1 - u^{alpha + 1}}{sqrt{1 - u^2}} , du, quad alpha neq -1.$$
    Enforcing a substitution of $u mapsto sqrt{u}$ results in
    $$I = frac{1}{2(alpha + 1)} int_0^1 left (frac{1}{sqrt{u(1 - u)}} - frac{u^{alpha/2}}{sqrt{ 1 - u}} right ) , du = I_1 - I_2.$$



    The first of the integrals is trivial
    $$I_1 = frac{1}{2(alpha + 1)} int_0^1 frac{du}{sqrt{frac{1}{4} - (u - frac{1}{2})^2}} = frac{1}{2(alpha + 1)} arcsin (2u - 1) Big{|}^1_0 = frac{pi}{2(alpha + 1)}.$$



    For the second of the integrals, it can be evaluated by writing it in terms of the beta function. Here
    begin{align}
    I_2 &= int_0^1 frac{u^{alpha/2}}{sqrt{1 - u}} , du\
    &= int_0^1 u^{(alpha/2 + 1) - 1} (1 - u)^{1/2 - 1} , du\
    &= text{B} left (frac{alpha}{2} + 1, frac{1}{2} right )\
    &= sqrt{pi} , frac{Gamma left (frac{alpha + 2}{2} right )}{Gamma left (frac{alpha + 3}{2} right )}.\
    end{align}
    Thus
    $$I = frac{1}{2(alpha + 1)} left [pi - sqrt{pi} , frac{Gamma left (frac{alpha + 2}{2} right )}{Gamma left (frac{alpha + 3}{2} right )} right ], qquad alpha neq -1.$$



    For the case when $alpha = -1$, the double integral at ($*$) becomes
    $$I = int_0^1 int_u^1 frac{1}{xsqrt{1 - u^2}} , dx du.$$
    After performing the $x$-integral which yields a natural logarithm, one has
    $$I = -int_0^1 frac{ln u}{sqrt{1 - u^2}} , du.$$
    Enforcing a substitution of $u mapsto sin u$ leads to
    $$I = -int_0^{pi/2} ln (sin u) , du. qquad (**)$$
    The integral appearing in ($**$) is quite famous and reasonably (?) well known. Its evaluation can be found either here or here. Thus
    $$I = frac{pi}{2} ln 2, qquad alpha = -1.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      A fantastic solution. Thank you very much very. I'm been working on developing my general nasty integral skills and this mode of converting a single to double integral is great. Is this a form of Feynman's Trick? Regardless, again, fantastic solution.
      $endgroup$
      – DavidG
      Dec 10 '18 at 12:10














    2












    2








    2





    $begingroup$

    Here is a method that relies on using a double integral.



    Noting the integral converges for $alpha > -2$, recognising
    $$arcsin x = int_0^x frac{du}{sqrt{1 - u^2}},$$
    the integral can be rewritten as
    $$I = int_0^1 int_0^x frac{x^alpha}{sqrt{1 - u^2}} , du dx.$$
    On changing the order of integration, one has
    begin{equation}
    int_0^1 int_u^1 frac{x^alpha}{sqrt{1 - u^2}} , dx du. qquad (*)
    end{equation}
    After performing the $x$-integral we are left with
    $$I = frac{1}{alpha + 1} int_0^1 frac{1 - u^{alpha + 1}}{sqrt{1 - u^2}} , du, quad alpha neq -1.$$
    Enforcing a substitution of $u mapsto sqrt{u}$ results in
    $$I = frac{1}{2(alpha + 1)} int_0^1 left (frac{1}{sqrt{u(1 - u)}} - frac{u^{alpha/2}}{sqrt{ 1 - u}} right ) , du = I_1 - I_2.$$



    The first of the integrals is trivial
    $$I_1 = frac{1}{2(alpha + 1)} int_0^1 frac{du}{sqrt{frac{1}{4} - (u - frac{1}{2})^2}} = frac{1}{2(alpha + 1)} arcsin (2u - 1) Big{|}^1_0 = frac{pi}{2(alpha + 1)}.$$



    For the second of the integrals, it can be evaluated by writing it in terms of the beta function. Here
    begin{align}
    I_2 &= int_0^1 frac{u^{alpha/2}}{sqrt{1 - u}} , du\
    &= int_0^1 u^{(alpha/2 + 1) - 1} (1 - u)^{1/2 - 1} , du\
    &= text{B} left (frac{alpha}{2} + 1, frac{1}{2} right )\
    &= sqrt{pi} , frac{Gamma left (frac{alpha + 2}{2} right )}{Gamma left (frac{alpha + 3}{2} right )}.\
    end{align}
    Thus
    $$I = frac{1}{2(alpha + 1)} left [pi - sqrt{pi} , frac{Gamma left (frac{alpha + 2}{2} right )}{Gamma left (frac{alpha + 3}{2} right )} right ], qquad alpha neq -1.$$



    For the case when $alpha = -1$, the double integral at ($*$) becomes
    $$I = int_0^1 int_u^1 frac{1}{xsqrt{1 - u^2}} , dx du.$$
    After performing the $x$-integral which yields a natural logarithm, one has
    $$I = -int_0^1 frac{ln u}{sqrt{1 - u^2}} , du.$$
    Enforcing a substitution of $u mapsto sin u$ leads to
    $$I = -int_0^{pi/2} ln (sin u) , du. qquad (**)$$
    The integral appearing in ($**$) is quite famous and reasonably (?) well known. Its evaluation can be found either here or here. Thus
    $$I = frac{pi}{2} ln 2, qquad alpha = -1.$$






    share|cite|improve this answer









    $endgroup$



    Here is a method that relies on using a double integral.



    Noting the integral converges for $alpha > -2$, recognising
    $$arcsin x = int_0^x frac{du}{sqrt{1 - u^2}},$$
    the integral can be rewritten as
    $$I = int_0^1 int_0^x frac{x^alpha}{sqrt{1 - u^2}} , du dx.$$
    On changing the order of integration, one has
    begin{equation}
    int_0^1 int_u^1 frac{x^alpha}{sqrt{1 - u^2}} , dx du. qquad (*)
    end{equation}
    After performing the $x$-integral we are left with
    $$I = frac{1}{alpha + 1} int_0^1 frac{1 - u^{alpha + 1}}{sqrt{1 - u^2}} , du, quad alpha neq -1.$$
    Enforcing a substitution of $u mapsto sqrt{u}$ results in
    $$I = frac{1}{2(alpha + 1)} int_0^1 left (frac{1}{sqrt{u(1 - u)}} - frac{u^{alpha/2}}{sqrt{ 1 - u}} right ) , du = I_1 - I_2.$$



    The first of the integrals is trivial
    $$I_1 = frac{1}{2(alpha + 1)} int_0^1 frac{du}{sqrt{frac{1}{4} - (u - frac{1}{2})^2}} = frac{1}{2(alpha + 1)} arcsin (2u - 1) Big{|}^1_0 = frac{pi}{2(alpha + 1)}.$$



    For the second of the integrals, it can be evaluated by writing it in terms of the beta function. Here
    begin{align}
    I_2 &= int_0^1 frac{u^{alpha/2}}{sqrt{1 - u}} , du\
    &= int_0^1 u^{(alpha/2 + 1) - 1} (1 - u)^{1/2 - 1} , du\
    &= text{B} left (frac{alpha}{2} + 1, frac{1}{2} right )\
    &= sqrt{pi} , frac{Gamma left (frac{alpha + 2}{2} right )}{Gamma left (frac{alpha + 3}{2} right )}.\
    end{align}
    Thus
    $$I = frac{1}{2(alpha + 1)} left [pi - sqrt{pi} , frac{Gamma left (frac{alpha + 2}{2} right )}{Gamma left (frac{alpha + 3}{2} right )} right ], qquad alpha neq -1.$$



    For the case when $alpha = -1$, the double integral at ($*$) becomes
    $$I = int_0^1 int_u^1 frac{1}{xsqrt{1 - u^2}} , dx du.$$
    After performing the $x$-integral which yields a natural logarithm, one has
    $$I = -int_0^1 frac{ln u}{sqrt{1 - u^2}} , du.$$
    Enforcing a substitution of $u mapsto sin u$ leads to
    $$I = -int_0^{pi/2} ln (sin u) , du. qquad (**)$$
    The integral appearing in ($**$) is quite famous and reasonably (?) well known. Its evaluation can be found either here or here. Thus
    $$I = frac{pi}{2} ln 2, qquad alpha = -1.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 10 '18 at 9:01









    omegadotomegadot

    6,4342829




    6,4342829












    • $begingroup$
      A fantastic solution. Thank you very much very. I'm been working on developing my general nasty integral skills and this mode of converting a single to double integral is great. Is this a form of Feynman's Trick? Regardless, again, fantastic solution.
      $endgroup$
      – DavidG
      Dec 10 '18 at 12:10


















    • $begingroup$
      A fantastic solution. Thank you very much very. I'm been working on developing my general nasty integral skills and this mode of converting a single to double integral is great. Is this a form of Feynman's Trick? Regardless, again, fantastic solution.
      $endgroup$
      – DavidG
      Dec 10 '18 at 12:10
















    $begingroup$
    A fantastic solution. Thank you very much very. I'm been working on developing my general nasty integral skills and this mode of converting a single to double integral is great. Is this a form of Feynman's Trick? Regardless, again, fantastic solution.
    $endgroup$
    – DavidG
    Dec 10 '18 at 12:10




    $begingroup$
    A fantastic solution. Thank you very much very. I'm been working on developing my general nasty integral skills and this mode of converting a single to double integral is great. Is this a form of Feynman's Trick? Regardless, again, fantastic solution.
    $endgroup$
    – DavidG
    Dec 10 '18 at 12:10











    1












    $begingroup$

    Answer 2.0:



    We know that for $|x|<1$,
    $$arcsin x=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)}x^{2k+1}$$
    Where $$(1/2)_k=frac{Gamma(1/2+k)}{Gamma(1/2)}$$
    Hence we may begin with
    $$F(a)=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)}int_0^1x^{2k+a+1}mathrm dx=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)(2k+2+a)}$$
    Then we play with the fractions a little to get
    $$F(a)=frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}bigg[frac1{2k+1}-frac1{2k+2+a}bigg]$$
    $$F(a)=frac1{a+1}sum_{ngeq0}frac{(1/2)_n}{n!}frac1{2n+1}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}frac1{2k+2+a}$$
    $$F(a)=fracpi{2(a+1)}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!(2k+2+a)}$$
    $$F(a)=fracpi{2(a+1)}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}int_0^1x^{2k+1+a}mathrm dx$$
    $$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1sum_{kgeq0}frac{(1/2)_k}{k!}x^{2k+1+a}mathrm dx$$
    $$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1frac{x^{a+1}}{Gamma(1/2)}sum_{kgeq0}frac{Gamma(1/2+k)}{k!}x^{2k}mathrm dx$$
    $$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1frac{x^{a+1}}{sqrt{1-x^2}}mathrm dx$$
    $u=x^2$:
    $$F(a)=fracpi{2(a+1)}-frac1{2(a+1)}int_0^1u^{a/2}(1-u)^{-1/2}mathrm du$$
    $$F(a)=fracpi{2(a+1)}-frac{Gamma(frac{a+2}2)Gamma(frac12)}{2(a+1)Gamma(frac{a+3}2)}$$
    $$F(a)=frac{sqrt{pi}}{2(a+1)}bigg[sqrt{pi}-frac{Gamma(frac{a+2}2)}{Gamma(frac{a+3}2)}bigg]$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Answer 2.0:



      We know that for $|x|<1$,
      $$arcsin x=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)}x^{2k+1}$$
      Where $$(1/2)_k=frac{Gamma(1/2+k)}{Gamma(1/2)}$$
      Hence we may begin with
      $$F(a)=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)}int_0^1x^{2k+a+1}mathrm dx=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)(2k+2+a)}$$
      Then we play with the fractions a little to get
      $$F(a)=frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}bigg[frac1{2k+1}-frac1{2k+2+a}bigg]$$
      $$F(a)=frac1{a+1}sum_{ngeq0}frac{(1/2)_n}{n!}frac1{2n+1}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}frac1{2k+2+a}$$
      $$F(a)=fracpi{2(a+1)}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!(2k+2+a)}$$
      $$F(a)=fracpi{2(a+1)}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}int_0^1x^{2k+1+a}mathrm dx$$
      $$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1sum_{kgeq0}frac{(1/2)_k}{k!}x^{2k+1+a}mathrm dx$$
      $$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1frac{x^{a+1}}{Gamma(1/2)}sum_{kgeq0}frac{Gamma(1/2+k)}{k!}x^{2k}mathrm dx$$
      $$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1frac{x^{a+1}}{sqrt{1-x^2}}mathrm dx$$
      $u=x^2$:
      $$F(a)=fracpi{2(a+1)}-frac1{2(a+1)}int_0^1u^{a/2}(1-u)^{-1/2}mathrm du$$
      $$F(a)=fracpi{2(a+1)}-frac{Gamma(frac{a+2}2)Gamma(frac12)}{2(a+1)Gamma(frac{a+3}2)}$$
      $$F(a)=frac{sqrt{pi}}{2(a+1)}bigg[sqrt{pi}-frac{Gamma(frac{a+2}2)}{Gamma(frac{a+3}2)}bigg]$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Answer 2.0:



        We know that for $|x|<1$,
        $$arcsin x=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)}x^{2k+1}$$
        Where $$(1/2)_k=frac{Gamma(1/2+k)}{Gamma(1/2)}$$
        Hence we may begin with
        $$F(a)=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)}int_0^1x^{2k+a+1}mathrm dx=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)(2k+2+a)}$$
        Then we play with the fractions a little to get
        $$F(a)=frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}bigg[frac1{2k+1}-frac1{2k+2+a}bigg]$$
        $$F(a)=frac1{a+1}sum_{ngeq0}frac{(1/2)_n}{n!}frac1{2n+1}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}frac1{2k+2+a}$$
        $$F(a)=fracpi{2(a+1)}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!(2k+2+a)}$$
        $$F(a)=fracpi{2(a+1)}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}int_0^1x^{2k+1+a}mathrm dx$$
        $$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1sum_{kgeq0}frac{(1/2)_k}{k!}x^{2k+1+a}mathrm dx$$
        $$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1frac{x^{a+1}}{Gamma(1/2)}sum_{kgeq0}frac{Gamma(1/2+k)}{k!}x^{2k}mathrm dx$$
        $$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1frac{x^{a+1}}{sqrt{1-x^2}}mathrm dx$$
        $u=x^2$:
        $$F(a)=fracpi{2(a+1)}-frac1{2(a+1)}int_0^1u^{a/2}(1-u)^{-1/2}mathrm du$$
        $$F(a)=fracpi{2(a+1)}-frac{Gamma(frac{a+2}2)Gamma(frac12)}{2(a+1)Gamma(frac{a+3}2)}$$
        $$F(a)=frac{sqrt{pi}}{2(a+1)}bigg[sqrt{pi}-frac{Gamma(frac{a+2}2)}{Gamma(frac{a+3}2)}bigg]$$






        share|cite|improve this answer









        $endgroup$



        Answer 2.0:



        We know that for $|x|<1$,
        $$arcsin x=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)}x^{2k+1}$$
        Where $$(1/2)_k=frac{Gamma(1/2+k)}{Gamma(1/2)}$$
        Hence we may begin with
        $$F(a)=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)}int_0^1x^{2k+a+1}mathrm dx=sum_{kgeq0}frac{(1/2)_k}{k!(2k+1)(2k+2+a)}$$
        Then we play with the fractions a little to get
        $$F(a)=frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}bigg[frac1{2k+1}-frac1{2k+2+a}bigg]$$
        $$F(a)=frac1{a+1}sum_{ngeq0}frac{(1/2)_n}{n!}frac1{2n+1}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}frac1{2k+2+a}$$
        $$F(a)=fracpi{2(a+1)}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!(2k+2+a)}$$
        $$F(a)=fracpi{2(a+1)}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}int_0^1x^{2k+1+a}mathrm dx$$
        $$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1sum_{kgeq0}frac{(1/2)_k}{k!}x^{2k+1+a}mathrm dx$$
        $$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1frac{x^{a+1}}{Gamma(1/2)}sum_{kgeq0}frac{Gamma(1/2+k)}{k!}x^{2k}mathrm dx$$
        $$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1frac{x^{a+1}}{sqrt{1-x^2}}mathrm dx$$
        $u=x^2$:
        $$F(a)=fracpi{2(a+1)}-frac1{2(a+1)}int_0^1u^{a/2}(1-u)^{-1/2}mathrm du$$
        $$F(a)=fracpi{2(a+1)}-frac{Gamma(frac{a+2}2)Gamma(frac12)}{2(a+1)Gamma(frac{a+3}2)}$$
        $$F(a)=frac{sqrt{pi}}{2(a+1)}bigg[sqrt{pi}-frac{Gamma(frac{a+2}2)}{Gamma(frac{a+3}2)}bigg]$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 17:28









        clathratusclathratus

        5,1701338




        5,1701338






























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            Can I use Tabulator js library in my java Spring + Thymeleaf project?

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            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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