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

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.

Answer 1

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.