# Where does $\Gamma \left( \frac{3}{2} \right) = \int _0 ^{+\infty}\! \mathrm e ^{-x^2} \, \mathrm d x$ come from?

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.

• 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 Feb 2, 2019 at 12:08
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 $$\int_0^\infty (x^s-sx^{s-1})\exp (-x) dx=[-x^s\exp (-x)]_0^\infty=0$$.