Integration of powers of the $sin x$












12












$begingroup$


I have to evalute
$$int_0^{frac{pi}{2}}(sin x)^z dx.$$
I put this integral in Wolfram Alpha, and the result is
$$frac{sqrt{pi}Gammaleft(frac{z+1}{2}right)}{2Gammaleft(frac{z}{2}+1right)},$$
but I don't know why. If $z$ is a positive integer, then one can do integration by parts, many times. Eventually, this yields
$$int_0^{frac{pi}{2}}(sin x)^{2z} dx=frac{(2z-1)!!}{(2z)!!}frac{pi}{2},$$
where $(2n-1)!!=1cdot 3cdots (2n-1)$, and $(2n)!!=2cdot 4cdots 2n$.
I appreciate your help.






integration gamma-function







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Did you look at the Wikipedia page on the Gamma-function? Everything you need for solving this problem is there.
    $endgroup$
    – t.b.
    Jul 9 '11 at 3:53






  • 2




    $begingroup$
    Beta-function...
    $endgroup$
    – Andrew
    Jul 9 '11 at 5:41










  • $begingroup$
    Yeah, of course. That is the thing. Thank you :)
    $endgroup$
    – leo
    Jul 9 '11 at 5:51










  • $begingroup$
    This is too easy for to be a post. Should I delete this post? What do you think @Theo and @Andrew?
    $endgroup$
    – leo
    Jul 9 '11 at 5:58








  • 1




    $begingroup$
    @leo: I don't think you should delete it. Why don't you write up your own solution and post them as an answer? If you ping me, I'll have a look.
    $endgroup$
    – t.b.
    Jul 9 '11 at 8:26
















12












$begingroup$


I have to evalute
$$int_0^{frac{pi}{2}}(sin x)^z dx.$$
I put this integral in Wolfram Alpha, and the result is
$$frac{sqrt{pi}Gammaleft(frac{z+1}{2}right)}{2Gammaleft(frac{z}{2}+1right)},$$
but I don't know why. If $z$ is a positive integer, then one can do integration by parts, many times. Eventually, this yields
$$int_0^{frac{pi}{2}}(sin x)^{2z} dx=frac{(2z-1)!!}{(2z)!!}frac{pi}{2},$$
where $(2n-1)!!=1cdot 3cdots (2n-1)$, and $(2n)!!=2cdot 4cdots 2n$.
I appreciate your help.






integration gamma-function







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Did you look at the Wikipedia page on the Gamma-function? Everything you need for solving this problem is there.
    $endgroup$
    – t.b.
    Jul 9 '11 at 3:53






  • 2




    $begingroup$
    Beta-function...
    $endgroup$
    – Andrew
    Jul 9 '11 at 5:41










  • $begingroup$
    Yeah, of course. That is the thing. Thank you :)
    $endgroup$
    – leo
    Jul 9 '11 at 5:51










  • $begingroup$
    This is too easy for to be a post. Should I delete this post? What do you think @Theo and @Andrew?
    $endgroup$
    – leo
    Jul 9 '11 at 5:58








  • 1




    $begingroup$
    @leo: I don't think you should delete it. Why don't you write up your own solution and post them as an answer? If you ping me, I'll have a look.
    $endgroup$
    – t.b.
    Jul 9 '11 at 8:26














12












12








12


6



$begingroup$


I have to evalute
$$int_0^{frac{pi}{2}}(sin x)^z dx.$$
I put this integral in Wolfram Alpha, and the result is
$$frac{sqrt{pi}Gammaleft(frac{z+1}{2}right)}{2Gammaleft(frac{z}{2}+1right)},$$
but I don't know why. If $z$ is a positive integer, then one can do integration by parts, many times. Eventually, this yields
$$int_0^{frac{pi}{2}}(sin x)^{2z} dx=frac{(2z-1)!!}{(2z)!!}frac{pi}{2},$$
where $(2n-1)!!=1cdot 3cdots (2n-1)$, and $(2n)!!=2cdot 4cdots 2n$.
I appreciate your help.






integration gamma-function







share|cite|improve this question











$endgroup$




I have to evalute
$$int_0^{frac{pi}{2}}(sin x)^z dx.$$
I put this integral in Wolfram Alpha, and the result is
$$frac{sqrt{pi}Gammaleft(frac{z+1}{2}right)}{2Gammaleft(frac{z}{2}+1right)},$$
but I don't know why. If $z$ is a positive integer, then one can do integration by parts, many times. Eventually, this yields
$$int_0^{frac{pi}{2}}(sin x)^{2z} dx=frac{(2z-1)!!}{(2z)!!}frac{pi}{2},$$
where $(2n-1)!!=1cdot 3cdots (2n-1)$, and $(2n)!!=2cdot 4cdots 2n$.
I appreciate your help.






integration gamma-function




integration gamma-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 11 '11 at 6:41







leo

















asked Jul 9 '11 at 3:14









leoleo

6,04553582




6,04553582








  • 6




    $begingroup$
    Did you look at the Wikipedia page on the Gamma-function? Everything you need for solving this problem is there.
    $endgroup$
    – t.b.
    Jul 9 '11 at 3:53






  • 2




    $begingroup$
    Beta-function...
    $endgroup$
    – Andrew
    Jul 9 '11 at 5:41










  • $begingroup$
    Yeah, of course. That is the thing. Thank you :)
    $endgroup$
    – leo
    Jul 9 '11 at 5:51










  • $begingroup$
    This is too easy for to be a post. Should I delete this post? What do you think @Theo and @Andrew?
    $endgroup$
    – leo
    Jul 9 '11 at 5:58








  • 1




    $begingroup$
    @leo: I don't think you should delete it. Why don't you write up your own solution and post them as an answer? If you ping me, I'll have a look.
    $endgroup$
    – t.b.
    Jul 9 '11 at 8:26














  • 6




    $begingroup$
    Did you look at the Wikipedia page on the Gamma-function? Everything you need for solving this problem is there.
    $endgroup$
    – t.b.
    Jul 9 '11 at 3:53






  • 2




    $begingroup$
    Beta-function...
    $endgroup$
    – Andrew
    Jul 9 '11 at 5:41










  • $begingroup$
    Yeah, of course. That is the thing. Thank you :)
    $endgroup$
    – leo
    Jul 9 '11 at 5:51










  • $begingroup$
    This is too easy for to be a post. Should I delete this post? What do you think @Theo and @Andrew?
    $endgroup$
    – leo
    Jul 9 '11 at 5:58








  • 1




    $begingroup$
    @leo: I don't think you should delete it. Why don't you write up your own solution and post them as an answer? If you ping me, I'll have a look.
    $endgroup$
    – t.b.
    Jul 9 '11 at 8:26








6




6




$begingroup$
Did you look at the Wikipedia page on the Gamma-function? Everything you need for solving this problem is there.
$endgroup$
– t.b.
Jul 9 '11 at 3:53




$begingroup$
Did you look at the Wikipedia page on the Gamma-function? Everything you need for solving this problem is there.
$endgroup$
– t.b.
Jul 9 '11 at 3:53




2




2




$begingroup$
Beta-function...
$endgroup$
– Andrew
Jul 9 '11 at 5:41




$begingroup$
Beta-function...
$endgroup$
– Andrew
Jul 9 '11 at 5:41












$begingroup$
Yeah, of course. That is the thing. Thank you :)
$endgroup$
– leo
Jul 9 '11 at 5:51




$begingroup$
Yeah, of course. That is the thing. Thank you :)
$endgroup$
– leo
Jul 9 '11 at 5:51












$begingroup$
This is too easy for to be a post. Should I delete this post? What do you think @Theo and @Andrew?
$endgroup$
– leo
Jul 9 '11 at 5:58






$begingroup$
This is too easy for to be a post. Should I delete this post? What do you think @Theo and @Andrew?
$endgroup$
– leo
Jul 9 '11 at 5:58






1




1




$begingroup$
@leo: I don't think you should delete it. Why don't you write up your own solution and post them as an answer? If you ping me, I'll have a look.
$endgroup$
– t.b.
Jul 9 '11 at 8:26




$begingroup$
@leo: I don't think you should delete it. Why don't you write up your own solution and post them as an answer? If you ping me, I'll have a look.
$endgroup$
– t.b.
Jul 9 '11 at 8:26










3 Answers
3






active

oldest

votes


















13












$begingroup$

The answer provided by Leo is the first one that comes to mind, but here is one starting directly from the definition of $Gamma(s)$.



From the definition of Gamma:



Consider



$$Gamma(s)Gamma(z)=int_{0}^{infty}int_{0}^{infty}t^{s-1}u^{z-1}e^{-(t+u)}dtdu.$$



Let $t=x^{2}$, $u=y^{2}$. Then we have



$$Gamma(s)Gamma(z)=4int_{0}^{infty}int_{0}^{infty}x^{2s-1}y^{2z-1}e^{-(x^{2}+y^{2})}dtdu.$$



Change to polar coordinates and set $y=rsintheta$, $x=rcostheta$, to get



$$Gamma(s)Gamma(z)=4left(int_{0}^{pi/2}cos^{2s-1}thetasin^{2z-1}theta dthetaright)left(int_{0}^{infty}r^{2s+2z-1}e^{-r^{2}}drright).$$



Letting $eta=r^{2}$ we get



$$2int_{0}^{infty}r^{2s+2z-1}e^{-r^{2}}dr=int_{0}^{infty}eta^{s+z-1}e^{-eta}deta=Gamma(s+z).$$



Hence



$$frac{Gamma(s)Gamma(z)}{Gamma(s+z)}=2left(int_{0}^{pi/2}cos^{2s-1}theta sin^{2z-1}theta dthetaright).$$



Setting $s=frac{1}{2}$ and $z=frac{x+1}{2}$ then yields your identity.



Hope that helps,






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's nice, as usual when you're manipulating these functions!
    $endgroup$
    – t.b.
    Jul 10 '11 at 1:39










  • $begingroup$
    Nice one @Eric.
    $endgroup$
    – leo
    Jul 10 '11 at 6:05



















2












$begingroup$

Just, following Theo's hint
$$
begin{align*}
int_{0}^{frac{pi}{2}}{(sinpsi)^x}dpsi&= int_{0}^{frac{pi}{2}}{(sinpsi)^{2cdot frac{1}{2}(x+1)-1}(cospsi)^{2cdot frac{1}{2}-1}}dpsi\
&=frac{1}{2}Bleft( frac{x+1}{2},frac{1}{2} right)\
&= frac{1}{2}cdot frac{Gammaleft(frac{x+1}{2}right)Gammaleft(frac{1}{2}right)}{Gammaleft( frac{x}{2}+1 right)}\
&=frac{sqrt{pi}Gammaleft(frac{x+1}{2}right)}{2Gammaleft( frac{x}{2}+1 right)}.
end{align*}$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Well, that's it (assuming that you understood all the identities you used). Also, have you verified that this is what you already knew in the case where $x$ is an integer?
    $endgroup$
    – t.b.
    Jul 10 '11 at 0:24












  • $begingroup$
    yep, and thaks @Theo.
    $endgroup$
    – leo
    Jul 10 '11 at 6:05



















0












$begingroup$

Use n instead of z. when n is even integer it is equal to 1.3.5....(2n-1)pi/2.4.6....n.(2) and when n is negetive 2.4.6....n pi/1.3.5....(2n-1).(2). Do u know how to get this result from this integral?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Welcome to MSE. Your answer adds nothing new to the already existing answers.
    $endgroup$
    – José Carlos Santos
    Jan 16 at 8:31










  • $begingroup$
    nice efforts! Go ahead to newer questions.
    $endgroup$
    – idea
    Jan 16 at 10:52











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f50447%2fintegration-of-powers-of-the-sin-x%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









13












$begingroup$

The answer provided by Leo is the first one that comes to mind, but here is one starting directly from the definition of $Gamma(s)$.



From the definition of Gamma:



Consider



$$Gamma(s)Gamma(z)=int_{0}^{infty}int_{0}^{infty}t^{s-1}u^{z-1}e^{-(t+u)}dtdu.$$



Let $t=x^{2}$, $u=y^{2}$. Then we have



$$Gamma(s)Gamma(z)=4int_{0}^{infty}int_{0}^{infty}x^{2s-1}y^{2z-1}e^{-(x^{2}+y^{2})}dtdu.$$



Change to polar coordinates and set $y=rsintheta$, $x=rcostheta$, to get



$$Gamma(s)Gamma(z)=4left(int_{0}^{pi/2}cos^{2s-1}thetasin^{2z-1}theta dthetaright)left(int_{0}^{infty}r^{2s+2z-1}e^{-r^{2}}drright).$$



Letting $eta=r^{2}$ we get



$$2int_{0}^{infty}r^{2s+2z-1}e^{-r^{2}}dr=int_{0}^{infty}eta^{s+z-1}e^{-eta}deta=Gamma(s+z).$$



Hence



$$frac{Gamma(s)Gamma(z)}{Gamma(s+z)}=2left(int_{0}^{pi/2}cos^{2s-1}theta sin^{2z-1}theta dthetaright).$$



Setting $s=frac{1}{2}$ and $z=frac{x+1}{2}$ then yields your identity.



Hope that helps,






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's nice, as usual when you're manipulating these functions!
    $endgroup$
    – t.b.
    Jul 10 '11 at 1:39










  • $begingroup$
    Nice one @Eric.
    $endgroup$
    – leo
    Jul 10 '11 at 6:05
















13












$begingroup$

The answer provided by Leo is the first one that comes to mind, but here is one starting directly from the definition of $Gamma(s)$.



From the definition of Gamma:



Consider



$$Gamma(s)Gamma(z)=int_{0}^{infty}int_{0}^{infty}t^{s-1}u^{z-1}e^{-(t+u)}dtdu.$$



Let $t=x^{2}$, $u=y^{2}$. Then we have



$$Gamma(s)Gamma(z)=4int_{0}^{infty}int_{0}^{infty}x^{2s-1}y^{2z-1}e^{-(x^{2}+y^{2})}dtdu.$$



Change to polar coordinates and set $y=rsintheta$, $x=rcostheta$, to get



$$Gamma(s)Gamma(z)=4left(int_{0}^{pi/2}cos^{2s-1}thetasin^{2z-1}theta dthetaright)left(int_{0}^{infty}r^{2s+2z-1}e^{-r^{2}}drright).$$



Letting $eta=r^{2}$ we get



$$2int_{0}^{infty}r^{2s+2z-1}e^{-r^{2}}dr=int_{0}^{infty}eta^{s+z-1}e^{-eta}deta=Gamma(s+z).$$



Hence



$$frac{Gamma(s)Gamma(z)}{Gamma(s+z)}=2left(int_{0}^{pi/2}cos^{2s-1}theta sin^{2z-1}theta dthetaright).$$



Setting $s=frac{1}{2}$ and $z=frac{x+1}{2}$ then yields your identity.



Hope that helps,






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's nice, as usual when you're manipulating these functions!
    $endgroup$
    – t.b.
    Jul 10 '11 at 1:39










  • $begingroup$
    Nice one @Eric.
    $endgroup$
    – leo
    Jul 10 '11 at 6:05














13












13








13





$begingroup$

The answer provided by Leo is the first one that comes to mind, but here is one starting directly from the definition of $Gamma(s)$.



From the definition of Gamma:



Consider



$$Gamma(s)Gamma(z)=int_{0}^{infty}int_{0}^{infty}t^{s-1}u^{z-1}e^{-(t+u)}dtdu.$$



Let $t=x^{2}$, $u=y^{2}$. Then we have



$$Gamma(s)Gamma(z)=4int_{0}^{infty}int_{0}^{infty}x^{2s-1}y^{2z-1}e^{-(x^{2}+y^{2})}dtdu.$$



Change to polar coordinates and set $y=rsintheta$, $x=rcostheta$, to get



$$Gamma(s)Gamma(z)=4left(int_{0}^{pi/2}cos^{2s-1}thetasin^{2z-1}theta dthetaright)left(int_{0}^{infty}r^{2s+2z-1}e^{-r^{2}}drright).$$



Letting $eta=r^{2}$ we get



$$2int_{0}^{infty}r^{2s+2z-1}e^{-r^{2}}dr=int_{0}^{infty}eta^{s+z-1}e^{-eta}deta=Gamma(s+z).$$



Hence



$$frac{Gamma(s)Gamma(z)}{Gamma(s+z)}=2left(int_{0}^{pi/2}cos^{2s-1}theta sin^{2z-1}theta dthetaright).$$



Setting $s=frac{1}{2}$ and $z=frac{x+1}{2}$ then yields your identity.



Hope that helps,






share|cite|improve this answer











$endgroup$



The answer provided by Leo is the first one that comes to mind, but here is one starting directly from the definition of $Gamma(s)$.



From the definition of Gamma:



Consider



$$Gamma(s)Gamma(z)=int_{0}^{infty}int_{0}^{infty}t^{s-1}u^{z-1}e^{-(t+u)}dtdu.$$



Let $t=x^{2}$, $u=y^{2}$. Then we have



$$Gamma(s)Gamma(z)=4int_{0}^{infty}int_{0}^{infty}x^{2s-1}y^{2z-1}e^{-(x^{2}+y^{2})}dtdu.$$



Change to polar coordinates and set $y=rsintheta$, $x=rcostheta$, to get



$$Gamma(s)Gamma(z)=4left(int_{0}^{pi/2}cos^{2s-1}thetasin^{2z-1}theta dthetaright)left(int_{0}^{infty}r^{2s+2z-1}e^{-r^{2}}drright).$$



Letting $eta=r^{2}$ we get



$$2int_{0}^{infty}r^{2s+2z-1}e^{-r^{2}}dr=int_{0}^{infty}eta^{s+z-1}e^{-eta}deta=Gamma(s+z).$$



Hence



$$frac{Gamma(s)Gamma(z)}{Gamma(s+z)}=2left(int_{0}^{pi/2}cos^{2s-1}theta sin^{2z-1}theta dthetaright).$$



Setting $s=frac{1}{2}$ and $z=frac{x+1}{2}$ then yields your identity.



Hope that helps,







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jul 10 '11 at 0:55

























answered Jul 10 '11 at 0:43









Eric NaslundEric Naslund

60.3k10139241




60.3k10139241












  • $begingroup$
    That's nice, as usual when you're manipulating these functions!
    $endgroup$
    – t.b.
    Jul 10 '11 at 1:39










  • $begingroup$
    Nice one @Eric.
    $endgroup$
    – leo
    Jul 10 '11 at 6:05


















  • $begingroup$
    That's nice, as usual when you're manipulating these functions!
    $endgroup$
    – t.b.
    Jul 10 '11 at 1:39










  • $begingroup$
    Nice one @Eric.
    $endgroup$
    – leo
    Jul 10 '11 at 6:05
















$begingroup$
That's nice, as usual when you're manipulating these functions!
$endgroup$
– t.b.
Jul 10 '11 at 1:39




$begingroup$
That's nice, as usual when you're manipulating these functions!
$endgroup$
– t.b.
Jul 10 '11 at 1:39












$begingroup$
Nice one @Eric.
$endgroup$
– leo
Jul 10 '11 at 6:05




$begingroup$
Nice one @Eric.
$endgroup$
– leo
Jul 10 '11 at 6:05











2












$begingroup$

Just, following Theo's hint
$$
begin{align*}
int_{0}^{frac{pi}{2}}{(sinpsi)^x}dpsi&= int_{0}^{frac{pi}{2}}{(sinpsi)^{2cdot frac{1}{2}(x+1)-1}(cospsi)^{2cdot frac{1}{2}-1}}dpsi\
&=frac{1}{2}Bleft( frac{x+1}{2},frac{1}{2} right)\
&= frac{1}{2}cdot frac{Gammaleft(frac{x+1}{2}right)Gammaleft(frac{1}{2}right)}{Gammaleft( frac{x}{2}+1 right)}\
&=frac{sqrt{pi}Gammaleft(frac{x+1}{2}right)}{2Gammaleft( frac{x}{2}+1 right)}.
end{align*}$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Well, that's it (assuming that you understood all the identities you used). Also, have you verified that this is what you already knew in the case where $x$ is an integer?
    $endgroup$
    – t.b.
    Jul 10 '11 at 0:24












  • $begingroup$
    yep, and thaks @Theo.
    $endgroup$
    – leo
    Jul 10 '11 at 6:05
















2












$begingroup$

Just, following Theo's hint
$$
begin{align*}
int_{0}^{frac{pi}{2}}{(sinpsi)^x}dpsi&= int_{0}^{frac{pi}{2}}{(sinpsi)^{2cdot frac{1}{2}(x+1)-1}(cospsi)^{2cdot frac{1}{2}-1}}dpsi\
&=frac{1}{2}Bleft( frac{x+1}{2},frac{1}{2} right)\
&= frac{1}{2}cdot frac{Gammaleft(frac{x+1}{2}right)Gammaleft(frac{1}{2}right)}{Gammaleft( frac{x}{2}+1 right)}\
&=frac{sqrt{pi}Gammaleft(frac{x+1}{2}right)}{2Gammaleft( frac{x}{2}+1 right)}.
end{align*}$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Well, that's it (assuming that you understood all the identities you used). Also, have you verified that this is what you already knew in the case where $x$ is an integer?
    $endgroup$
    – t.b.
    Jul 10 '11 at 0:24












  • $begingroup$
    yep, and thaks @Theo.
    $endgroup$
    – leo
    Jul 10 '11 at 6:05














2












2








2





$begingroup$

Just, following Theo's hint
$$
begin{align*}
int_{0}^{frac{pi}{2}}{(sinpsi)^x}dpsi&= int_{0}^{frac{pi}{2}}{(sinpsi)^{2cdot frac{1}{2}(x+1)-1}(cospsi)^{2cdot frac{1}{2}-1}}dpsi\
&=frac{1}{2}Bleft( frac{x+1}{2},frac{1}{2} right)\
&= frac{1}{2}cdot frac{Gammaleft(frac{x+1}{2}right)Gammaleft(frac{1}{2}right)}{Gammaleft( frac{x}{2}+1 right)}\
&=frac{sqrt{pi}Gammaleft(frac{x+1}{2}right)}{2Gammaleft( frac{x}{2}+1 right)}.
end{align*}$$






share|cite|improve this answer











$endgroup$



Just, following Theo's hint
$$
begin{align*}
int_{0}^{frac{pi}{2}}{(sinpsi)^x}dpsi&= int_{0}^{frac{pi}{2}}{(sinpsi)^{2cdot frac{1}{2}(x+1)-1}(cospsi)^{2cdot frac{1}{2}-1}}dpsi\
&=frac{1}{2}Bleft( frac{x+1}{2},frac{1}{2} right)\
&= frac{1}{2}cdot frac{Gammaleft(frac{x+1}{2}right)Gammaleft(frac{1}{2}right)}{Gammaleft( frac{x}{2}+1 right)}\
&=frac{sqrt{pi}Gammaleft(frac{x+1}{2}right)}{2Gammaleft( frac{x}{2}+1 right)}.
end{align*}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jul 1 '15 at 4:27

























answered Jul 10 '11 at 0:17









leoleo

6,04553582




6,04553582








  • 1




    $begingroup$
    Well, that's it (assuming that you understood all the identities you used). Also, have you verified that this is what you already knew in the case where $x$ is an integer?
    $endgroup$
    – t.b.
    Jul 10 '11 at 0:24












  • $begingroup$
    yep, and thaks @Theo.
    $endgroup$
    – leo
    Jul 10 '11 at 6:05














  • 1




    $begingroup$
    Well, that's it (assuming that you understood all the identities you used). Also, have you verified that this is what you already knew in the case where $x$ is an integer?
    $endgroup$
    – t.b.
    Jul 10 '11 at 0:24












  • $begingroup$
    yep, and thaks @Theo.
    $endgroup$
    – leo
    Jul 10 '11 at 6:05








1




1




$begingroup$
Well, that's it (assuming that you understood all the identities you used). Also, have you verified that this is what you already knew in the case where $x$ is an integer?
$endgroup$
– t.b.
Jul 10 '11 at 0:24






$begingroup$
Well, that's it (assuming that you understood all the identities you used). Also, have you verified that this is what you already knew in the case where $x$ is an integer?
$endgroup$
– t.b.
Jul 10 '11 at 0:24














$begingroup$
yep, and thaks @Theo.
$endgroup$
– leo
Jul 10 '11 at 6:05




$begingroup$
yep, and thaks @Theo.
$endgroup$
– leo
Jul 10 '11 at 6:05











0












$begingroup$

Use n instead of z. when n is even integer it is equal to 1.3.5....(2n-1)pi/2.4.6....n.(2) and when n is negetive 2.4.6....n pi/1.3.5....(2n-1).(2). Do u know how to get this result from this integral?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Welcome to MSE. Your answer adds nothing new to the already existing answers.
    $endgroup$
    – José Carlos Santos
    Jan 16 at 8:31










  • $begingroup$
    nice efforts! Go ahead to newer questions.
    $endgroup$
    – idea
    Jan 16 at 10:52
















0












$begingroup$

Use n instead of z. when n is even integer it is equal to 1.3.5....(2n-1)pi/2.4.6....n.(2) and when n is negetive 2.4.6....n pi/1.3.5....(2n-1).(2). Do u know how to get this result from this integral?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Welcome to MSE. Your answer adds nothing new to the already existing answers.
    $endgroup$
    – José Carlos Santos
    Jan 16 at 8:31










  • $begingroup$
    nice efforts! Go ahead to newer questions.
    $endgroup$
    – idea
    Jan 16 at 10:52














0












0








0





$begingroup$

Use n instead of z. when n is even integer it is equal to 1.3.5....(2n-1)pi/2.4.6....n.(2) and when n is negetive 2.4.6....n pi/1.3.5....(2n-1).(2). Do u know how to get this result from this integral?






share|cite|improve this answer









$endgroup$



Use n instead of z. when n is even integer it is equal to 1.3.5....(2n-1)pi/2.4.6....n.(2) and when n is negetive 2.4.6....n pi/1.3.5....(2n-1).(2). Do u know how to get this result from this integral?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 16 at 8:15









S M NaserS M Naser

1




1












  • $begingroup$
    Welcome to MSE. Your answer adds nothing new to the already existing answers.
    $endgroup$
    – José Carlos Santos
    Jan 16 at 8:31










  • $begingroup$
    nice efforts! Go ahead to newer questions.
    $endgroup$
    – idea
    Jan 16 at 10:52


















  • $begingroup$
    Welcome to MSE. Your answer adds nothing new to the already existing answers.
    $endgroup$
    – José Carlos Santos
    Jan 16 at 8:31










  • $begingroup$
    nice efforts! Go ahead to newer questions.
    $endgroup$
    – idea
    Jan 16 at 10:52
















$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Jan 16 at 8:31




$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Jan 16 at 8:31












$begingroup$
nice efforts! Go ahead to newer questions.
$endgroup$
– idea
Jan 16 at 10:52




$begingroup$
nice efforts! Go ahead to newer questions.
$endgroup$
– idea
Jan 16 at 10:52


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f50447%2fintegration-of-powers-of-the-sin-x%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
Read more

The Binding of Isaac: Rebirth/Afterbirth

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
Read more