Cfrgtkky

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.

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
$$I = frac{pi}{2} ln 2, qquad alpha = -1.$$

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
$$F(a)=frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}bigg[frac1{2k+1}-frac1{2k+2+a}bigg]$$
$$F(a)=frac1{a+1}sum_{ngeq0}frac{(1/2)_n}{n!}frac1{2n+1}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}frac1{2k+2+a}$$
$$F(a)=fracpi{2(a+1)}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!(2k+2+a)}$$
$$F(a)=fracpi{2(a+1)}-frac1{a+1}sum_{kgeq0}frac{(1/2)_k}{k!}int_0^1x^{2k+1+a}mathrm dx$$
$$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1sum_{kgeq0}frac{(1/2)_k}{k!}x^{2k+1+a}mathrm dx$$
$$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1frac{x^{a+1}}{Gamma(1/2)}sum_{kgeq0}frac{Gamma(1/2+k)}{k!}x^{2k}mathrm dx$$
$$F(a)=fracpi{2(a+1)}-frac1{a+1}int_0^1frac{x^{a+1}}{sqrt{1-x^2}}mathrm dx$$
$u=x^2$:
$$F(a)=fracpi{2(a+1)}-frac1{2(a+1)}int_0^1u^{a/2}(1-u)^{-1/2}mathrm du$$
$$F(a)=fracpi{2(a+1)}-frac{Gamma(frac{a+2}2)Gamma(frac12)}{2(a+1)Gamma(frac{a+3}2)}$$
$$F(a)=frac{sqrt{pi}}{2(a+1)}bigg[sqrt{pi}-frac{Gamma(frac{a+2}2)}{Gamma(frac{a+3}2)}bigg]$$