TITLE: Math competition problem, prove that $\int_{-\infty}^\infty e^{-\pi x^2 \left(\frac{\alpha +x}{\beta +x}\right)^2}dx=1~$ for $~0<\beta<\alpha$.
QUESTION [10 upvotes]: Recently there was a math competition in our university where this question 

Question: Prove that $\displaystyle\int_{-\infty}^\infty e^{-\pi  x^2 \left(\frac{\scriptstyle\alpha +x}{\scriptstyle\beta +x}\right)^2}dx=1~$ for $~0<\beta<\alpha$

has been asked, but nobody could solve it. I know that
$$
\int_{-\infty}^\infty e^{-\pi  x^2}dx=1
$$
but this doesn't help much. What are possible routes to deal with this kind of integrals? Any integration experts has any clue how this is done? Thanks.

REPLY [10 votes]: One may write
$$
 x \left(\frac{\alpha +x}{\beta +x}\right)=x-\frac{(\alpha-\beta)\beta}{x+\beta}+\alpha-\beta \tag1
$$ then, since $(\alpha-\beta)\beta>0$, one may use the G. Boole (1857) result,

$$
\int_{-\infty}^{+\infty}f\left(x-\frac{a}{x-b}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x,\qquad a>0.  \tag2
$$ 

giving, with $(1)$,
$$
\begin{align}
\int_{-\infty}^\infty e^{-\pi  x^2 \left(\frac{\scriptstyle\alpha +x}{\scriptstyle\beta +x}\right)^2}dx&=\int_{-\infty}^\infty e^{-\pi \left(x-\frac{(\alpha-\beta)\beta}{x+\beta}+\alpha-\beta \right)^2}dx
\\\\&=\int_{-\infty}^\infty e^{-\pi \left(x+\alpha-\beta \right)^2}dx
\\\\&=\int_{-\infty}^\infty e^{-\pi x^2}dx
\\\\&=1
\end{align}
$$
as announced.