# Using the Euler-Poisson integral, prove the identity

We have: $$I(a) =\int_{-\infty}^{+\infty}e^{-(ax^2+2bx)}dx$$ To prove: $$I(a) = \sqrt{\frac\pi a}e^{b^2/a}$$

I tried to differencate both sides, and got this:

Left side:

$$I'(a) = -2a\int_{-\infty}^{+\infty}xe^{-(ax^2+2bx)}dx -I(a)$$

Right side:

$$I'(a) = -I(a)\frac{b^2}{a^2}-I(a)\frac{1}{2a}$$

• How about the substitution $y=\sqrt a(x+b/a)$? Sep 17, 2019 at 10:39

Use $$y=x+b/a$$ so $$I(a)=\int_{\Bbb R}e^{-ay^2+b^2/a}dy$$, so the problem reduces to proving $$\int_{\Bbb R}e^{-ay^2}dy=\sqrt{\frac{\pi}{a}}$$. You can do the rest yourself (there are many ways to get $$I(1)$$, which is the crux of it).
notice that: $$ax^2+2bx=\left(\sqrt{a}x+\frac{b}{\sqrt{a}}\right)^2-\frac{b^2}{a}$$ Then you can make the substitution $$u=\sqrt{a}x+\frac{b}{\sqrt{a}}$$ and it can then be easily defined using well known definitions