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.

Have you tried contour integration? I feel like that might work. — mathworker21, Dec 17 '16 at 19:06
Just an idea: your starting point might actually be helpful if you do it the "Feynman way": for $\beta = \alpha$ the integrand reduces to your simpler special case. Now take the partial derivative with respect to $\beta$ (hopefully constant). — Max Freiburghaus, Dec 17 '16 at 19:17
Yes Glasser's theorem does solve this, but it's a bit too heavy machinery to be used on a competition problem imo. Would be interesting to know if this is in the curriculum / supposed to be known for students taking this problem. — Winther, Dec 17 '16 at 20:14
Similar type of question which might be useful: Is $\int_{\mathbb R} f(\sum_{k=1}^n\frac{1}{x-x_k})dx$ independent of $x_k$'s for certain $f$? The answer seems to provide an answer for this question also. — Winther, Dec 17 '16 at 20:20
@winther i guess that in this case a more elementary solution exists. I can't believe that Glasser is in the standard cirriculum at any university at the planet — tired, Dec 17 '16 at 21:49

score 10 · Answer 1 · answered Dec 17 '16 at 19:48

10

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.

answered Dec 17 '16 at 19:48

Olivier Oloa

120,989

It's always you :) – JPhy Dec 17 '16 at 21:30
I can't except this.There must be a more elementary solution. – Tyrell Dec 27 '16 at 19:45
@Tyrell I gave the general result, but in our case this is just an elementary change of variable, please see this: http://math.stackexchange.com/questions/457231/how-to-prove-int-infty-infty-fxdx-int-infty-infty-f-left/1461393#1461393 – Olivier Oloa Jan 01 '17 at 00:34

Math competition problem, prove that $\int_{-\infty}^\infty e^{-\pi x^2 \left(\frac{\alpha +x}{\beta +x}\right)^2}dx=1~$ for $~0<\beta<\alpha$.

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