Yesterday, during reviewing my old lecture notes on advanced quantum mechanics, i stumbeled over the following integral identity, which seems, on a first glance, too nice to be true

$$ I_{A,B}=\int_0^1\frac{1}{\sqrt{y^3(1-y)}}\exp\left(\frac{i A}{y}+\frac{i B}{1-y}\right)dy=\sqrt{\frac{i\pi}{B }}e^{i(\sqrt{A}+\sqrt{B})^2} $$

with $A,B>0$

After working on it for a few hours i came up with a solution, which i think is a bit cumbersome and will be presented below.

My questions to the community are the following:

What other solutions exists for this wonderful problem?

Especially is it possible to solve this integral by contour integration?


$\underline{\text{My solution:}}$

Start with $y=x^2$, now $$ I_{A,B}=2\int_0^{1}\frac{1}{x^2\sqrt{1-x^2}}\exp\left(\frac{i A}{x^2}+\frac{i B}{1-x^2}\right)dx $$ Next $x=\sin(q)$ and we get, $$ I_{A,B}=2\int_0^{\pi/2}\frac{1}{\sin^2(q)}\exp\left(\frac{i A}{\sin(q)^2}+\frac{i B}{\cos(q)^2}\right)dq$$

Simple trigonometric manipulations yield

$$ I_{A,B}=2\int_0^{\pi/2} e^{iA (\text{cot}(q)^2+1)+iB (\tan(q)^2+1)}\left(\text{cot}(q)^2+1\right)dq $$

Now the magic happens: $q=\text{arccot(p)}$

$$ I_{A,B}=2e^{i(A+B)}\int_0^{\infty}e^{i A p^2+i \frac{B}{p^2}}dp $$

This last integral is solvable by different methods..

$$ I_{A,B}=2e^{i(A+B)}\times \sqrt{\frac{i\pi}{4 B}}e^{i2\sqrt{AB}}=\sqrt{\frac{i\pi}{B }}e^{i(\sqrt{A}+\sqrt{B})^2} $$


  • 1
    $\begingroup$ Note that this is a convolution integral and may be solved by considering the Laplace transform of each individual function. math.stackexchange.com/questions/1148493/… $\endgroup$ – Ron Gordon Oct 7 '16 at 16:59
  • $\begingroup$ @RonGordon clever idea...(+1). Do think my question is too close to a duplicate of the linked one...maybe i should delete it $\endgroup$ – tired Oct 7 '16 at 17:04
  • $\begingroup$ No, it is not a duplicate. It is along the same lines. But don't delete it. $\endgroup$ – Ron Gordon Oct 7 '16 at 17:05
  • $\begingroup$ @RonGordon ok, it is interesting to note that my method also answer the question u answered back then $\endgroup$ – tired Oct 7 '16 at 17:07
  • $\begingroup$ Yes, yes it does. Or should. $\endgroup$ – Ron Gordon Oct 7 '16 at 17:08

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