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.
Best Answer
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 $\int_0^\infty (x^s-sx^{s-1})\exp (-x) dx=[-x^s\exp (-x)]_0^\infty=0$.