Mixing
Where does $Gamma left( frac{3}{2} right) = int _0 ^{+infty}! mathrm e ^{-x^2} , mathrm d x$ come from?
$begingroup$
My teacher solved this problem in class but I don't get how one step is justified.
Prove that $$int_0 ^{+infty} ! mathrm e ^{- x^2 } , mathrm d x = dfrac{sqrt{pi}}{2}$$ using this relation $$int_0 ^{+infty} ! int_0^{+infty} ! y ,mathrm e ^{-(1+ x^2 )y} , mathrm d y , mathrm d x = dfrac{pi}{4}.$$
Using Fubini's theorem we switch integrals:
$$int_0 ^{+infty} ! y, mathrm e ^{-y} left( int_0 ^{+infty} ! mathrm e ^{-x^2 y} , mathrm d x right) mathrm d y = dfrac{pi}{4}.$$
Let us compute first:
$$int_0 ^{+infty} ! mathrm e ^{-x^2 y} , mathrm d x =int _0 ^{+infty} ! dfrac{mathrm e ^{-t ^2}}{sqrt y}, mathrm d t= dfrac{mathcal E}{sqrt y},$$
where we have made the change $xsqrt y =t $ and $mathcal E$ is the integral that we want to compute.
Then $$int _0 ^{+infty} ! y , mathrm e ^{-y} dfrac{mathcal E}{sqrt y} , mathrm d y = mathcal E int _0 ^{+infty} y ^{frac{1}{2}} , mathrm e ^{-y} , mathrm d y = $$
$$=mathcal E int _0 ^{+infty} y ^{frac{3}{2}-1} , mathrm e ^{-y} , mathrm d y = mathcal E , Gamma left( frac{3}{2} right) color{red}{stackrel{?}{=}} $$
$$color{red}{stackrel{?}{=}} mathcal E int_0 ^{+infty} mathrm e ^{-s ^2} , mathrm d s = mathcal E ^2 = dfrac{pi}{4},$$
therefore $$mathcal E = dfrac{sqrt pi}{2} = int _0 ^{+infty}! mathrm e ^{-x^2} , mathrm d x.$$
What I don't get is how does he relate $mathcal E$ with the gamma function, that is, $$Gamma left( frac{3}{2} right) = int _0 ^{+infty}! mathrm e ^{-x^2} , mathrm d x = mathcal E.$$
I have seen that $Gamma left( frac{3}{2} right) = dfrac{sqrt pi }{2}$, but since we don't know the value of $mathcal E$ yet (as this is what we are trying to prove), this is not a way to relate them.
Thank you for your help.
integration improper-integrals gamma-function
asked Feb 2 at 11:36
mkspkmkspk
588617
$endgroup$
add a comment |
$begingroup$
My teacher solved this problem in class but I don't get how one step is justified.
Prove that $$int_0 ^{+infty} ! mathrm e ^{- x^2 } , mathrm d x = dfrac{sqrt{pi}}{2}$$ using this relation $$int_0 ^{+infty} ! int_0^{+infty} ! y ,mathrm e ^{-(1+ x^2 )y} , mathrm d y , mathrm d x = dfrac{pi}{4}.$$
Using Fubini's theorem we switch integrals:
$$int_0 ^{+infty} ! y, mathrm e ^{-y} left( int_0 ^{+infty} ! mathrm e ^{-x^2 y} , mathrm d x right) mathrm d y = dfrac{pi}{4}.$$
Let us compute first:
$$int_0 ^{+infty} ! mathrm e ^{-x^2 y} , mathrm d x =int _0 ^{+infty} ! dfrac{mathrm e ^{-t ^2}}{sqrt y}, mathrm d t= dfrac{mathcal E}{sqrt y},$$
where we have made the change $xsqrt y =t $ and $mathcal E$ is the integral that we want to compute.
Then $$int _0 ^{+infty} ! y , mathrm e ^{-y} dfrac{mathcal E}{sqrt y} , mathrm d y = mathcal E int _0 ^{+infty} y ^{frac{1}{2}} , mathrm e ^{-y} , mathrm d y = $$
$$=mathcal E int _0 ^{+infty} y ^{frac{3}{2}-1} , mathrm e ^{-y} , mathrm d y = mathcal E , Gamma left( frac{3}{2} right) color{red}{stackrel{?}{=}} $$
$$color{red}{stackrel{?}{=}} mathcal E int_0 ^{+infty} mathrm e ^{-s ^2} , mathrm d s = mathcal E ^2 = dfrac{pi}{4},$$
therefore $$mathcal E = dfrac{sqrt pi}{2} = int _0 ^{+infty}! mathrm e ^{-x^2} , mathrm d x.$$
What I don't get is how does he relate $mathcal E$ with the gamma function, that is, $$Gamma left( frac{3}{2} right) = int _0 ^{+infty}! mathrm e ^{-x^2} , mathrm d x = mathcal E.$$
I have seen that $Gamma left( frac{3}{2} right) = dfrac{sqrt pi }{2}$, but since we don't know the value of $mathcal E$ yet (as this is what we are trying to prove), this is not a way to relate them.
Thank you for your help.
integration improper-integrals gamma-function
asked Feb 2 at 11:36
mkspkmkspk
588617
$endgroup$
3
$begingroup$
I don't get why the downvote. Maybe I have explained it in a bad way, but this question has context and my research about my question got me to the conclusion that my teacher proved something using the thing he is trying to prove (and I am asking if there is something I am missing).
$endgroup$
– mkspk
Feb 2 at 11:46
2
$begingroup$
Just leave the step $Gammabig(frac{3}{2}big) = mathcal{E}$ and use what you know $Gammabig(frac{3}{2}big) = frac{sqrt{pi}}{2}$. Then your calculation leads to $$mathcal{E} frac{sqrt{pi}}{2} = frac{pi}{4},$$ which also solves your problem. By the way this also solves your questioned equality
$endgroup$
– Nathanael Skrepek
Feb 2 at 12:08
add a comment |
$begingroup$
My teacher solved this problem in class but I don't get how one step is justified.
Prove that $$int_0 ^{+infty} ! mathrm e ^{- x^2 } , mathrm d x = dfrac{sqrt{pi}}{2}$$ using this relation $$int_0 ^{+infty} ! int_0^{+infty} ! y ,mathrm e ^{-(1+ x^2 )y} , mathrm d y , mathrm d x = dfrac{pi}{4}.$$
Using Fubini's theorem we switch integrals:
$$int_0 ^{+infty} ! y, mathrm e ^{-y} left( int_0 ^{+infty} ! mathrm e ^{-x^2 y} , mathrm d x right) mathrm d y = dfrac{pi}{4}.$$
Let us compute first:
$$int_0 ^{+infty} ! mathrm e ^{-x^2 y} , mathrm d x =int _0 ^{+infty} ! dfrac{mathrm e ^{-t ^2}}{sqrt y}, mathrm d t= dfrac{mathcal E}{sqrt y},$$
where we have made the change $xsqrt y =t $ and $mathcal E$ is the integral that we want to compute.
Then $$int _0 ^{+infty} ! y , mathrm e ^{-y} dfrac{mathcal E}{sqrt y} , mathrm d y = mathcal E int _0 ^{+infty} y ^{frac{1}{2}} , mathrm e ^{-y} , mathrm d y = $$
$$=mathcal E int _0 ^{+infty} y ^{frac{3}{2}-1} , mathrm e ^{-y} , mathrm d y = mathcal E , Gamma left( frac{3}{2} right) color{red}{stackrel{?}{=}} $$
$$color{red}{stackrel{?}{=}} mathcal E int_0 ^{+infty} mathrm e ^{-s ^2} , mathrm d s = mathcal E ^2 = dfrac{pi}{4},$$
therefore $$mathcal E = dfrac{sqrt pi}{2} = int _0 ^{+infty}! mathrm e ^{-x^2} , mathrm d x.$$
What I don't get is how does he relate $mathcal E$ with the gamma function, that is, $$Gamma left( frac{3}{2} right) = int _0 ^{+infty}! mathrm e ^{-x^2} , mathrm d x = mathcal E.$$
I have seen that $Gamma left( frac{3}{2} right) = dfrac{sqrt pi }{2}$, but since we don't know the value of $mathcal E$ yet (as this is what we are trying to prove), this is not a way to relate them.
Thank you for your help.
integration improper-integrals gamma-function
asked Feb 2 at 11:36
mkspkmkspk
588617
$endgroup$
My teacher solved this problem in class but I don't get how one step is justified.
Prove that $$int_0 ^{+infty} ! mathrm e ^{- x^2 } , mathrm d x = dfrac{sqrt{pi}}{2}$$ using this relation $$int_0 ^{+infty} ! int_0^{+infty} ! y ,mathrm e ^{-(1+ x^2 )y} , mathrm d y , mathrm d x = dfrac{pi}{4}.$$
Using Fubini's theorem we switch integrals:
$$int_0 ^{+infty} ! y, mathrm e ^{-y} left( int_0 ^{+infty} ! mathrm e ^{-x^2 y} , mathrm d x right) mathrm d y = dfrac{pi}{4}.$$
Let us compute first:
$$int_0 ^{+infty} ! mathrm e ^{-x^2 y} , mathrm d x =int _0 ^{+infty} ! dfrac{mathrm e ^{-t ^2}}{sqrt y}, mathrm d t= dfrac{mathcal E}{sqrt y},$$
where we have made the change $xsqrt y =t $ and $mathcal E$ is the integral that we want to compute.
Then $$int _0 ^{+infty} ! y , mathrm e ^{-y} dfrac{mathcal E}{sqrt y} , mathrm d y = mathcal E int _0 ^{+infty} y ^{frac{1}{2}} , mathrm e ^{-y} , mathrm d y = $$
$$=mathcal E int _0 ^{+infty} y ^{frac{3}{2}-1} , mathrm e ^{-y} , mathrm d y = mathcal E , Gamma left( frac{3}{2} right) color{red}{stackrel{?}{=}} $$
$$color{red}{stackrel{?}{=}} mathcal E int_0 ^{+infty} mathrm e ^{-s ^2} , mathrm d s = mathcal E ^2 = dfrac{pi}{4},$$
therefore $$mathcal E = dfrac{sqrt pi}{2} = int _0 ^{+infty}! mathrm e ^{-x^2} , mathrm d x.$$
What I don't get is how does he relate $mathcal E$ with the gamma function, that is, $$Gamma left( frac{3}{2} right) = int _0 ^{+infty}! mathrm e ^{-x^2} , mathrm d x = mathcal E.$$
I have seen that $Gamma left( frac{3}{2} right) = dfrac{sqrt pi }{2}$, but since we don't know the value of $mathcal E$ yet (as this is what we are trying to prove), this is not a way to relate them.
Thank you for your help.
integration improper-integrals gamma-function
integration improper-integrals gamma-function
asked Feb 2 at 11:36
mkspkmkspk
588617
asked Feb 2 at 11:36
mkspkmkspk
588617
edited Feb 2 at 11:49
mkspk
asked Feb 2 at 11:36
mkspkmkspk
588617
asked Feb 2 at 11:36
mkspkmkspk
588617
asked Feb 2 at 11:36
mkspkmkspk
588617
588617
3
$begingroup$
I don't get why the downvote. Maybe I have explained it in a bad way, but this question has context and my research about my question got me to the conclusion that my teacher proved something using the thing he is trying to prove (and I am asking if there is something I am missing).
$endgroup$
– mkspk
Feb 2 at 11:46
2
$begingroup$
Just leave the step $Gammabig(frac{3}{2}big) = mathcal{E}$ and use what you know $Gammabig(frac{3}{2}big) = frac{sqrt{pi}}{2}$. Then your calculation leads to $$mathcal{E} frac{sqrt{pi}}{2} = frac{pi}{4},$$ which also solves your problem. By the way this also solves your questioned equality
$endgroup$
– Nathanael Skrepek
Feb 2 at 12:08
add a comment |
3
$begingroup$
I don't get why the downvote. Maybe I have explained it in a bad way, but this question has context and my research about my question got me to the conclusion that my teacher proved something using the thing he is trying to prove (and I am asking if there is something I am missing).
$endgroup$
– mkspk
Feb 2 at 11:46
2
$begingroup$
Just leave the step $Gammabig(frac{3}{2}big) = mathcal{E}$ and use what you know $Gammabig(frac{3}{2}big) = frac{sqrt{pi}}{2}$. Then your calculation leads to $$mathcal{E} frac{sqrt{pi}}{2} = frac{pi}{4},$$ which also solves your problem. By the way this also solves your questioned equality
$endgroup$
– Nathanael Skrepek
Feb 2 at 12:08
3
3
$begingroup$
I don't get why the downvote. Maybe I have explained it in a bad way, but this question has context and my research about my question got me to the conclusion that my teacher proved something using the thing he is trying to prove (and I am asking if there is something I am missing).
$endgroup$
– mkspk
Feb 2 at 11:46
$begingroup$
I don't get why the downvote. Maybe I have explained it in a bad way, but this question has context and my research about my question got me to the conclusion that my teacher proved something using the thing he is trying to prove (and I am asking if there is something I am missing).
$endgroup$
– mkspk
Feb 2 at 11:46
2
2
$begingroup$
Just leave the step $Gammabig(frac{3}{2}big) = mathcal{E}$ and use what you know $Gammabig(frac{3}{2}big) = frac{sqrt{pi}}{2}$. Then your calculation leads to $$mathcal{E} frac{sqrt{pi}}{2} = frac{pi}{4},$$ which also solves your problem. By the way this also solves your questioned equality
$endgroup$
– Nathanael Skrepek
Feb 2 at 12:08
$begingroup$
Just leave the step $Gammabig(frac{3}{2}big) = mathcal{E}$ and use what you know $Gammabig(frac{3}{2}big) = frac{sqrt{pi}}{2}$. Then your calculation leads to $$mathcal{E} frac{sqrt{pi}}{2} = frac{pi}{4},$$ which also solves your problem. By the way this also solves your questioned equality
$endgroup$
– Nathanael Skrepek
Feb 2 at 12:08
add a comment |
1 Answer
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$begingroup$
One definition of the Gamma function is $Gamma(s)=int_0^infty x^{s-1}exp (-x) dx=2int_0^infty x^{2s-1}exp (-x^2) dx$, so your integral is $frac12Gamma(frac12)$. One only need then use the identity $Gamma(s+1)=sGamma(s)$. Indeed, the difference is $int_0^infty (x^s-sx^{s-1})exp (-x) dx=[-x^sexp (-x)]_0^infty=0$.
edited Feb 2 at 19:16
Botond
6,57531034
answered Feb 2 at 11:50
J.G.J.G.
33.3k23252
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add a comment |
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$begingroup$
One definition of the Gamma function is $Gamma(s)=int_0^infty x^{s-1}exp (-x) dx=2int_0^infty x^{2s-1}exp (-x^2) dx$, so your integral is $frac12Gamma(frac12)$. One only need then use the identity $Gamma(s+1)=sGamma(s)$. Indeed, the difference is $int_0^infty (x^s-sx^{s-1})exp (-x) dx=[-x^sexp (-x)]_0^infty=0$.
edited Feb 2 at 19:16
Botond
6,57531034
answered Feb 2 at 11:50
J.G.J.G.
33.3k23252
$endgroup$
add a comment |
$begingroup$
One definition of the Gamma function is $Gamma(s)=int_0^infty x^{s-1}exp (-x) dx=2int_0^infty x^{2s-1}exp (-x^2) dx$, so your integral is $frac12Gamma(frac12)$. One only need then use the identity $Gamma(s+1)=sGamma(s)$. Indeed, the difference is $int_0^infty (x^s-sx^{s-1})exp (-x) dx=[-x^sexp (-x)]_0^infty=0$.
edited Feb 2 at 19:16
Botond
6,57531034
answered Feb 2 at 11:50
J.G.J.G.
33.3k23252
$endgroup$
add a comment |
$begingroup$
One definition of the Gamma function is $Gamma(s)=int_0^infty x^{s-1}exp (-x) dx=2int_0^infty x^{2s-1}exp (-x^2) dx$, so your integral is $frac12Gamma(frac12)$. One only need then use the identity $Gamma(s+1)=sGamma(s)$. Indeed, the difference is $int_0^infty (x^s-sx^{s-1})exp (-x) dx=[-x^sexp (-x)]_0^infty=0$.
edited Feb 2 at 19:16
Botond
6,57531034
answered Feb 2 at 11:50
J.G.J.G.
33.3k23252
$endgroup$
One definition of the Gamma function is $Gamma(s)=int_0^infty x^{s-1}exp (-x) dx=2int_0^infty x^{2s-1}exp (-x^2) dx$, so your integral is $frac12Gamma(frac12)$. One only need then use the identity $Gamma(s+1)=sGamma(s)$. Indeed, the difference is $int_0^infty (x^s-sx^{s-1})exp (-x) dx=[-x^sexp (-x)]_0^infty=0$.
edited Feb 2 at 19:16
Botond
6,57531034
answered Feb 2 at 11:50
J.G.J.G.
33.3k23252
edited Feb 2 at 19:16
Botond
6,57531034
edited Feb 2 at 19:16
Botond
6,57531034
edited Feb 2 at 19:16
Botond
6,57531034
6,57531034
answered Feb 2 at 11:50
J.G.J.G.
33.3k23252
answered Feb 2 at 11:50
J.G.J.G.
33.3k23252
answered Feb 2 at 11:50
J.G.J.G.
33.3k23252
33.3k23252
add a comment |
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$begingroup$
I don't get why the downvote. Maybe I have explained it in a bad way, but this question has context and my research about my question got me to the conclusion that my teacher proved something using the thing he is trying to prove (and I am asking if there is something I am missing).
$endgroup$
– mkspk
Feb 2 at 11:46
2
$begingroup$
Just leave the step $Gammabig(frac{3}{2}big) = mathcal{E}$ and use what you know $Gammabig(frac{3}{2}big) = frac{sqrt{pi}}{2}$. Then your calculation leads to $$mathcal{E} frac{sqrt{pi}}{2} = frac{pi}{4},$$ which also solves your problem. By the way this also solves your questioned equality
$endgroup$
– Nathanael Skrepek
Feb 2 at 12:08