finding $int frac{(1-x^2)^{n-1}}{(1+x^2)^n}mathrm{d}x$
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 at 5:52
clathratus
2,852327
add a comment |
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 at 5:52
clathratus
2,852327
add a comment |
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 at 5:52
clathratus
2,852327
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 at 5:52
clathratus
2,852327
asked Nov 25 at 5:52
clathratus
2,852327
asked Nov 25 at 5:52
clathratus
2,852327
asked Nov 25 at 5:52
clathratus
2,852327
asked Nov 25 at 5:52
clathratus
2,852327
2,852327
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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 at 5:56
Nosrati
26.4k62353
I feel so stupid right now. Thanks (+1)
– clathratus
Nov 25 at 5:58
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012480%2ffinding-int-frac1-x2n-11x2n-mathrmdx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 at 5:56
Nosrati
26.4k62353
I feel so stupid right now. Thanks (+1)
– clathratus
Nov 25 at 5:58
add a comment |
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 at 5:56
Nosrati
26.4k62353
I feel so stupid right now. Thanks (+1)
– clathratus
Nov 25 at 5:58
add a comment |
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 at 5:56
Nosrati
26.4k62353
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 at 5:56
Nosrati
26.4k62353
answered Nov 25 at 5:56
Nosrati
26.4k62353
answered Nov 25 at 5:56
Nosrati
26.4k62353
answered Nov 25 at 5:56
Nosrati
26.4k62353
26.4k62353
I feel so stupid right now. Thanks (+1)
– clathratus
Nov 25 at 5:58
add a comment |
I feel so stupid right now. Thanks (+1)
– clathratus
Nov 25 at 5:58
I feel so stupid right now. Thanks (+1)
– clathratus
Nov 25 at 5:58
I feel so stupid right now. Thanks (+1)
– clathratus
Nov 25 at 5:58
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012480%2ffinding-int-frac1-x2n-11x2n-mathrmdx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
