-1
$\begingroup$

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:

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

$\endgroup$
1
  • 2
    $\begingroup$ How about the substitution $y=\sqrt a(x+b/a)$? $\endgroup$ Sep 17, 2019 at 10:39

2 Answers 2

2
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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

$\endgroup$

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.