4
$\begingroup$

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 $x\sqrt 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.

$\endgroup$
  • 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 $\Gamma\big(\frac{3}{2}\big) = \mathcal{E}$ and use what you know $\Gamma\big(\frac{3}{2}\big) = \frac{\sqrt{\pi}}{2}$. 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
2
$\begingroup$

One definition of the Gamma function is $\Gamma(s)=\int_0^\infty x^{s-1}\exp (-x) dx=2\int_0^\infty x^{2s-1}\exp (-x^2) dx$, so your integral is $\frac12\Gamma(\frac12)$. One only need then use the identity $\Gamma(s+1)=s\Gamma(s)$. Indeed, the difference is $\int_0^\infty (x^s-sx^{s-1})\exp (-x) dx=[-x^s\exp (-x)]_0^\infty=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.