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In a physical mathematical problem, I came across a nontrivial triple integral below obtained upon 3D inverse Fourier transformation. It would be great if someone here could provide with some hints that could help to evaluate analytically $$ I =\frac{1}{(2\pi)^3} \int_0^{2\pi} \int_0^\infty \int_0^{\pi} \frac{\sin\theta \sin^2\phi}{a \cos^2 \theta + b \sin^2 \theta} \, e^{ikh\sin\theta\cos\phi} \, \mathrm{d} \theta \, \mathrm{d} k \, \mathrm{d} \phi \, . $$ where $a$ and $b$ are two positive real numbers. Using the change of variable $q=\cos\theta$, the latter equation can be written as $$ I = \frac{1}{(2\pi)^3} \int_0^{2\pi} \int_0^\infty \int_{-1}^{1} \frac{\sin^2\phi}{(a-b)q^2+b} \, e^{ikh\sqrt{1-q^2}\cos\phi} \, \mathrm{d} q \, \mathrm{d} k \, \mathrm{d} \phi \, . $$

A guess solution using Maple for some numerical values for $a$ and $b$ is obtained as $$ I = \frac{1}{4\pi h \sqrt{ab}} \, . $$

But, is there a way to prove that really? Any idea / feedback is welcome.

Thanks a lot!

Fede

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Well, for the first integral we have:

$$\mathscr{I}:=\int_0^{2\pi}\int_0^\infty\int_0^\pi\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)\cdot\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\space\text{d}\theta\space\text{d}\text{k}\space\text{d}\phi\tag1$$

Using Fubini's theorem:

$$\mathscr{I}:=\int_0^{2\pi}\int_0^\pi\int_0^\infty\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)\cdot\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\space\text{d}\text{k}\space\text{d}\theta\space\text{d}\phi=$$ $$\int_0^{2\pi}\int_0^\pi\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\int_0^\infty\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)\space\text{d}\text{k}\space\text{d}\theta\space\text{d}\phi\tag2$$

Now, we can use:

$$\int_0^\infty\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)\space\text{d}\text{k}=\frac{i}{\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{h}}\tag3$$

when $\Im\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{h}\right)>0$

So, we get:

$$\mathscr{I}=\int_0^{2\pi}\int_0^\pi\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\cdot\frac{i}{\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{h}}\space\text{d}\theta\space\text{d}\phi=$$ $$\int_0^{2\pi}\frac{\sin^2\left(\phi\right)\cdot i}{\cos\left(\phi\right)\cdot\text{h}}\int_0^\pi\frac{\sin\left(\theta\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\cdot\frac{1}{\sin\left(\theta\right)}\space\text{d}\theta\space\text{d}\phi=$$ $$\int_0^{2\pi}\frac{\sin^2\left(\phi\right)\cdot i}{\cos\left(\phi\right)\cdot\text{h}}\int_0^\pi\frac{1}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\space\text{d}\theta\space\text{d}\phi\tag4$$

Now, use:

$$\int_0^\pi\frac{1}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\space\text{d}\theta=\frac{\pi}{\sqrt{\text{a}\cdot\text{b}}}\tag5$$

So, we get:

$$\mathscr{I}=\int_0^{2\pi}\frac{\sin^2\left(\phi\right)\cdot i}{\cos\left(\phi\right)\cdot\text{h}}\cdot\frac{\pi}{\sqrt{\text{a}\cdot\text{b}}}\space\text{d}\phi=\frac{\pi\cdot i}{\text{h}\cdot\sqrt{\text{a}\cdot\text{b}}}\int_0^{2\pi}\frac{\sin^2\left(\phi\right)}{\cos\left(\phi\right)}\space\text{d}\phi\tag6$$

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  • $\begingroup$ Thanks for your helpful answer. But PV of the last integral is identically null. Does it mean that $I= 0 $ ? thanks $\endgroup$ – Math Student Jul 9 '17 at 13:26
  • $\begingroup$ I think that it shouldn't. The integral has a physical interpretation and should be strictly positive defined. The guess solution has been proved to be true using Maple but I don't know how to prove it. Indeed, your solution shows the desired $h\sqrt{ab}$ in the denominator. I think that the correct solution should not be that far... $\endgroup$ – Math Student Jul 9 '17 at 14:10
  • $\begingroup$ Yes. I should prove it. It is part of my doctoral research.. $\endgroup$ – Math Student Jul 9 '17 at 14:15
  • $\begingroup$ The integral is obtained when I am working on particle motion in an anisotropic medium (physics) you may not be interested in that matter.. It would be cool if you could please be of help thanks $\endgroup$ – Math Student Jul 9 '17 at 14:19
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    $\begingroup$ @Federiko Notice that you've to find the integral in the right order, from the inside one to the ouside one not the other way around! $\endgroup$ – Jan Jul 9 '17 at 20:01

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