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







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$endgroup$








  • 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











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







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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


















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