finding $int frac{(1-x^2)^{n-1}}{(1+x^2)^n}mathrm{d}x$
$begingroup$
While trying to evaluate a different integral, I met the following integral:
$$I_n=intfrac{(1-x^2)^{n-1}}{(1+x^2)^n}mathrm{d}x$$
I was lost as to what to do with it, so I looked it up on wolfram alpha, and I got a result in terms of The Appell Hypergeometric Function $F_1$:
$$I_n=xF_1bigg(frac12;1-n,n;frac32;x^2,-x^2bigg)+C$$
Where, for $|x|<1$, and $|y|<1$,
$$F_1(a;b_1,b_2;c;x,y)=sum_{mgeq0}sum_{kgeq0}frac{(a)_{m+k}(b_1)_m(b_2)_k}{m!k!(c)_{m+k}}x^my^k$$
With $$(x)_k=frac{Gamma(x+k)}{Gamma(x)}$$
I found on Wikipedia that
$$F_1(a;b1,b2;c;x,y)=\sum_{kgeq0}frac{(a)_k(b_1)_k(b_2)_k(c-a)_k}{(c+k-1)_k(c)_{2k}}frac{x^ky^k}{k!},_2F_1(a+k,b_1+k;c+2k;x),_2F_1(a+k,b_2+k;c+2k;y)$$
Which I thought might be useful.
I also know that
$$,_1F_0(1-n;;x^2)=(1-x^2)^{n-1}$$
Which gives $$I_n=sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{x^{2m}}{(1+x^2)^n}mathrm{d}x$$
One could preform the substitution $u=1+x^2$:
$$I_n=frac12sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{(u-1)^{m-1/2}}{u^n}mathrm{d}u$$
But this doesn't look any better.
Could anyone provide an alternative solution or give a proof of the result given by Wolfram?
integration special-functions
asked Nov 25 '18 at 5:52
clathratusclathratus
3,596332
$endgroup$
add a comment |
$begingroup$
While trying to evaluate a different integral, I met the following integral:
$$I_n=intfrac{(1-x^2)^{n-1}}{(1+x^2)^n}mathrm{d}x$$
I was lost as to what to do with it, so I looked it up on wolfram alpha, and I got a result in terms of The Appell Hypergeometric Function $F_1$:
$$I_n=xF_1bigg(frac12;1-n,n;frac32;x^2,-x^2bigg)+C$$
Where, for $|x|<1$, and $|y|<1$,
$$F_1(a;b_1,b_2;c;x,y)=sum_{mgeq0}sum_{kgeq0}frac{(a)_{m+k}(b_1)_m(b_2)_k}{m!k!(c)_{m+k}}x^my^k$$
With $$(x)_k=frac{Gamma(x+k)}{Gamma(x)}$$
I found on Wikipedia that
$$F_1(a;b1,b2;c;x,y)=\sum_{kgeq0}frac{(a)_k(b_1)_k(b_2)_k(c-a)_k}{(c+k-1)_k(c)_{2k}}frac{x^ky^k}{k!},_2F_1(a+k,b_1+k;c+2k;x),_2F_1(a+k,b_2+k;c+2k;y)$$
Which I thought might be useful.
I also know that
$$,_1F_0(1-n;;x^2)=(1-x^2)^{n-1}$$
Which gives $$I_n=sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{x^{2m}}{(1+x^2)^n}mathrm{d}x$$
One could preform the substitution $u=1+x^2$:
$$I_n=frac12sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{(u-1)^{m-1/2}}{u^n}mathrm{d}u$$
But this doesn't look any better.
Could anyone provide an alternative solution or give a proof of the result given by Wolfram?
integration special-functions
asked Nov 25 '18 at 5:52
clathratusclathratus
3,596332
$endgroup$
add a comment |
$begingroup$
While trying to evaluate a different integral, I met the following integral:
$$I_n=intfrac{(1-x^2)^{n-1}}{(1+x^2)^n}mathrm{d}x$$
I was lost as to what to do with it, so I looked it up on wolfram alpha, and I got a result in terms of The Appell Hypergeometric Function $F_1$:
$$I_n=xF_1bigg(frac12;1-n,n;frac32;x^2,-x^2bigg)+C$$
Where, for $|x|<1$, and $|y|<1$,
$$F_1(a;b_1,b_2;c;x,y)=sum_{mgeq0}sum_{kgeq0}frac{(a)_{m+k}(b_1)_m(b_2)_k}{m!k!(c)_{m+k}}x^my^k$$
With $$(x)_k=frac{Gamma(x+k)}{Gamma(x)}$$
I found on Wikipedia that
$$F_1(a;b1,b2;c;x,y)=\sum_{kgeq0}frac{(a)_k(b_1)_k(b_2)_k(c-a)_k}{(c+k-1)_k(c)_{2k}}frac{x^ky^k}{k!},_2F_1(a+k,b_1+k;c+2k;x),_2F_1(a+k,b_2+k;c+2k;y)$$
Which I thought might be useful.
I also know that
$$,_1F_0(1-n;;x^2)=(1-x^2)^{n-1}$$
Which gives $$I_n=sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{x^{2m}}{(1+x^2)^n}mathrm{d}x$$
One could preform the substitution $u=1+x^2$:
$$I_n=frac12sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{(u-1)^{m-1/2}}{u^n}mathrm{d}u$$
But this doesn't look any better.
Could anyone provide an alternative solution or give a proof of the result given by Wolfram?
integration special-functions
asked Nov 25 '18 at 5:52
clathratusclathratus
3,596332
$endgroup$
While trying to evaluate a different integral, I met the following integral:
$$I_n=intfrac{(1-x^2)^{n-1}}{(1+x^2)^n}mathrm{d}x$$
I was lost as to what to do with it, so I looked it up on wolfram alpha, and I got a result in terms of The Appell Hypergeometric Function $F_1$:
$$I_n=xF_1bigg(frac12;1-n,n;frac32;x^2,-x^2bigg)+C$$
Where, for $|x|<1$, and $|y|<1$,
$$F_1(a;b_1,b_2;c;x,y)=sum_{mgeq0}sum_{kgeq0}frac{(a)_{m+k}(b_1)_m(b_2)_k}{m!k!(c)_{m+k}}x^my^k$$
With $$(x)_k=frac{Gamma(x+k)}{Gamma(x)}$$
I found on Wikipedia that
$$F_1(a;b1,b2;c;x,y)=\sum_{kgeq0}frac{(a)_k(b_1)_k(b_2)_k(c-a)_k}{(c+k-1)_k(c)_{2k}}frac{x^ky^k}{k!},_2F_1(a+k,b_1+k;c+2k;x),_2F_1(a+k,b_2+k;c+2k;y)$$
Which I thought might be useful.
I also know that
$$,_1F_0(1-n;;x^2)=(1-x^2)^{n-1}$$
Which gives $$I_n=sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{x^{2m}}{(1+x^2)^n}mathrm{d}x$$
One could preform the substitution $u=1+x^2$:
$$I_n=frac12sum_{mgeq0}frac{(1-n)_m}{m!}intfrac{(u-1)^{m-1/2}}{u^n}mathrm{d}u$$
But this doesn't look any better.
Could anyone provide an alternative solution or give a proof of the result given by Wolfram?
integration special-functions
integration special-functions
asked Nov 25 '18 at 5:52
clathratusclathratus
3,596332
asked Nov 25 '18 at 5:52
clathratusclathratus
3,596332
asked Nov 25 '18 at 5:52
clathratusclathratus
3,596332
asked Nov 25 '18 at 5:52
clathratusclathratus
3,596332
asked Nov 25 '18 at 5:52
clathratusclathratus
3,596332
3,596332
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Hint:
With substitution $x=tan t$ the integral is
$$I_n=int cos^{n-1}2t dt$$
and it can be solved with integration by parts and recursive formula.
answered Nov 25 '18 at 5:56
NosratiNosrati
26.5k62354
$endgroup$
$begingroup$
I feel so stupid right now. Thanks (+1)
$endgroup$
– clathratus
Nov 25 '18 at 5:58
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
With substitution $x=tan t$ the integral is
$$I_n=int cos^{n-1}2t dt$$
and it can be solved with integration by parts and recursive formula.
answered Nov 25 '18 at 5:56
NosratiNosrati
26.5k62354
$endgroup$
$begingroup$
I feel so stupid right now. Thanks (+1)
$endgroup$
– clathratus
Nov 25 '18 at 5:58
add a comment |
$begingroup$
Hint:
With substitution $x=tan t$ the integral is
$$I_n=int cos^{n-1}2t dt$$
and it can be solved with integration by parts and recursive formula.
answered Nov 25 '18 at 5:56
NosratiNosrati
26.5k62354
$endgroup$
$begingroup$
I feel so stupid right now. Thanks (+1)
$endgroup$
– clathratus
Nov 25 '18 at 5:58
add a comment |
$begingroup$
Hint:
With substitution $x=tan t$ the integral is
$$I_n=int cos^{n-1}2t dt$$
and it can be solved with integration by parts and recursive formula.
answered Nov 25 '18 at 5:56
NosratiNosrati
26.5k62354
$endgroup$
Hint:
With substitution $x=tan t$ the integral is
$$I_n=int cos^{n-1}2t dt$$
and it can be solved with integration by parts and recursive formula.
answered Nov 25 '18 at 5:56
NosratiNosrati
26.5k62354
answered Nov 25 '18 at 5:56
NosratiNosrati
26.5k62354
answered Nov 25 '18 at 5:56
NosratiNosrati
26.5k62354
answered Nov 25 '18 at 5:56
NosratiNosrati
26.5k62354
26.5k62354
$begingroup$
I feel so stupid right now. Thanks (+1)
$endgroup$
– clathratus
Nov 25 '18 at 5:58
add a comment |
$begingroup$
I feel so stupid right now. Thanks (+1)
$endgroup$
– clathratus
Nov 25 '18 at 5:58
$begingroup$
I feel so stupid right now. Thanks (+1)
$endgroup$
– clathratus
Nov 25 '18 at 5:58
$begingroup$
I feel so stupid right now. Thanks (+1)
$endgroup$
– clathratus
Nov 25 '18 at 5:58
add a comment |
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