Difficult trigonometric integral. [Solved]

I have faced some difficulties to do the following integral:

$$I=\int_{0}^{2\pi}d\phi\int_{0}^{\pi}d\theta~\sin\theta\int_{0}^{\infty}dr~r^2\frac{3x^2y^2\cos(u r \sin\theta \cos\phi)\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}\mathrm e^{-\frac{r^2}{2}} \tag{1},$$

where $x$, $y$, and $u$ are real positive constants. I tried at least two ways to solve this integral:

• First attempt:

I began to solve the $r$ integral first. By using Mathematica, then

$$I=\int_{0}^{2\pi}d\phi\int_{0}^{\pi}d\theta~\sin\theta\frac{3x^2y^2(1-u^2\sin^2\theta\cos^2\phi)\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}\mathrm e^{-\frac{u^2}{2}\sin^2\theta\cos^2\phi} \tag{2}.$$

After that, I looked for a solution for $\phi$ integral. My best attempt was:

$$I_\phi(x,y,u,\theta)=\frac{2}{B}\left[B\left(\frac{1}{2}\right)F_1\left(\frac{1}{2},1,-;1;\nu,-\frac{a}{2}\right)-aB\left(\frac{1}{2}\right)F_1\left(\frac{3}{2},1,-;2;\nu,-\frac{a}{2}\right)\right],$$

where $B=x^2\sin^2\theta+x^2y^2\cos^2\theta$, $a=u^2\sin^2\theta$, and $\nu=\frac{x^2-y^2}{x^2+x^2y^2\cot^2\theta}$. In this way, the final results it's something like that:

$$I= \int_{0}^{\pi} \mathrm d \theta~3x^2y^2\sin\theta \cos^2\theta~ I_\phi(x,y,u,\theta). \tag{3}.$$

Eq. $(3)$ cannot be further simplied in general and is the nal result.

• Second attempt:

To avoid the hypergeometric function $F_1$, I tried to start with the $\phi$ integral. In this case, my initial problem is an integral something like that:

$$\int_{0}^{2\pi} \mathrm d \phi \frac{\cos(A \cos\phi)}{a^2\cos^2\phi+b^2\sin^2\phi}. \tag{4}$$

This integral $(4)$ can be solved by series (see Vincent's answer and Jack's answer). However those solutions, at least for me, has not a closed form. This is my final step on this second attempt :(

What is the point? It turns out that someone has managed to solve the integral $(1)$, at least the integral in $r$ and $\phi$. The final resuls found by this person was:

$$I_G=\frac{12 \pi x~y}{(1-x^2)^{3/2}}\int_{0}^{\sqrt{1-x^2}} \mathrm dk \frac{k^2 \exp\left(-\frac{u^2}{2}\frac{x^2k^2}{(1-x^2)(1-k^2)}\right)}{\sqrt{1-k^2}\sqrt{1-k^2\frac{1-y^2}{1-x^2}}},$$

where, I belive, $k=\sqrt{1-x^2}\cos\theta$. As you can see in this following code performed in Mathematica

IG[x_, y_, u_] :=
Sqrt[Pi/2] NIntegrate[(12  Pi x y)/(1 - x^2)^(3/2)
(v^2 Exp[-(u^2 x^2 v^2)/(2 (1 - x^2) (1 - v^2))])/(Sqrt[1 - v^2] Sqrt[1 - v^2 (1 - y^2)/(1 - x^2)]), {v, 0, Sqrt[1 - x^2]}]
IG[.3, .4, 1]
** 4.53251 **

I[x_, y_, u_] :=
NIntegrate[(r^2 Sin[a] Cos[
u r Sin[a] Cos[b]] 3 x^2 y^2 Cos[a]^2 Exp[-r^2/
2])/((y^2 Cos[b]^2 + x^2 Sin[b]^2) Sin[a]^2 +
x^2 y^2 Cos[a]^2), {r, 0, Infinity}, {a, 0, Pi}, {b, 0, 2 Pi}]
I[.3, .4, 1]
** 4.53251 **


the integrals $I$ and $I_G$ are equals. Indeed, since that they emerge from the same physical problem.

So, my question is: what are the steps applied for that integral $I$ gives the integral $I_G$?

Edit

Since my question was not solved yet, I think it is because it is a tough question, I will show a particular case of the integral $I$, letting $u=0$. I hope with this help you help me.

In this case, the $r$ integral in $(1)$ is trivial and the integral takes the form:

$$I_P=\int_{0}^{2\pi}d\phi\int_{0}^{\pi}d\theta~\sin\theta\frac{3x^2y^2\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}. \tag{5}$$

The $\phi$ integral can be integrated with the help of Eq. 3.642.1 in Gradstein and Ryzhik's tables of integrals. Thereby, the $I_P$ takes the for:

$$I_P=3xy\int_{0}^{\pi}d\theta\frac{\sin\theta\cos^2\theta}{\sqrt{1+(x^2-1)\cos^2\theta}\sqrt{1+(y^2-1)\cos^2\theta}}. \tag{6}$$

Now the change of variable $k=\sqrt{1-x^2}\cos\theta$ bring expression $(6)$ to the form

$$I_P= \frac{(const) x~y}{(1-x^2)^{3/2}}\int_{0}^{\sqrt{1-x^2}} \mathrm dk \frac{k^2}{\sqrt{1-k^2}\sqrt{1-k^2\frac{1-y^2}{1-x^2}}}.$$

Did you notice how $I_G$ and $I_P$ are similar? Do you think a similar approach can be applied to my original problem? Please, let me know.

Edit 2

The integral $(1)$ is also evaluated in Appendix A.4 of this thesis. However, there he used cylindrical symmetry.

Edit: ended

My bounty ended, and unfortunately, I don't have enough reputation to offer another one. My question was not solved. Perhaps to solve that it is necessary to make some physical consideration. Anyway, I thanks to all who helped me. If I can solve this, I put the solution here.

I've solved this problem applying the Schwinger proper-time substitution: $$\frac{1}{q^2}=\int_{0}^{\infty}\mathrm{d\xi}~\mathrm{e^{-q^2\xi}}$$

• at this moment i don't think any of these can be evaluated in terms of elementary function!!!!! – user547564 Aug 19 '18 at 19:07
• But there is one. As soon as possible I will post the solution found by an acquaintance of mine, which is numerically equal to equation 1. – Dinesh Shankar Aug 20 '18 at 10:19
• When you say 'therefore, it is possible to find an elementary closed form', why do you believe that having a closed form in integrating from one direction implies that you're likely to have a closed form integrating from another? – Steven Stadnicki Aug 21 '18 at 14:52
• Perhaps link to the MO copy of the question is worth adding: I really don't know what else I can do to solve this integral. – Martin Sleziak Aug 21 '18 at 18:20
• Where's @Cleo when you need her (user profile)? :P – Mr Pie Aug 23 '18 at 1:15

By using the Maclaurin series of $\exp(x)$ and $\cos(x)$ both integrals boil down to the computation of

$$\int_{0}^{\pi/2}\frac{\left(\cos x\right)^n}{a^2 \sin^2(x)+b^2\cos^2(x)}\,dx=\int_{0}^{\pi/2}\frac{\left(\sin x\right)^n}{a^2 \sin^2(x)+b^2\cos^2(x)}\,dx$$ which can be tackled thrugh the tangent half-angle substitution and the residue theorem.

• Nice. Thank you. Mathematica can do it! – Dinesh Shankar Aug 20 '18 at 10:21
• Jack, unfortunately, this solution does not give the correct answer :( – Dinesh Shankar Aug 21 '18 at 10:03
• Why do I think this approach was not helpful to me? Because the result of this integral, according to Mathematica is: $$\frac{1}{2} \pi \sec \left(\frac{\pi n}{2}\right) \left(\frac{\left(\frac{b^2}{a^2}\right)^{\frac{n+1}{2}} \left(1-\frac{b^2}{a^2}\right)^{-n/2}}{b^2}-\frac{\sqrt{\pi } \, _2\tilde{F}_1\left(\frac{1}{2},1;\frac{3-n}{2};\frac{b^2}{a^2}\right)}{a^2 \Gamma \left(\frac{n}{2}\right)}\right)$$. Please let me know if I made a mistake. – Dinesh Shankar Aug 22 '18 at 9:20
• By summing the shown contributions you get a hypergeometric function, there is nothing wrong here. – Jack D'Aurizio Aug 22 '18 at 12:22
• @user3321 whether you made a mistake or not, that big math comment looks beautiful :P – Mr Pie Aug 23 '18 at 1:14

Here is an outline of the approach I have taken to solve this integral.

First rewrite the integral $(1)$ in Cartesian variables:

$$I=\int_{-\infty}^{\infty} \mathrm{d}^3v~ \frac{3x^2y^2v_z^2}{y^2v_x^2+x^2v_y^2+x^2y^2v_z^2}\cos(uv_x)\exp\left(-\frac{v_x^2}{2}-\frac{v_y^2}{2}-\frac{v_z^2}{2}\right).$$

Now use the following substitution

$$\frac{1}{y^2v_x^2+x^2v_y^2+x^2y^2v_z^2}=\int_{0}^{\infty}d\tau~\mathrm{ e^{-(y^2v_x^2+x^2v_y^2+x^2y^2v_z^2)\tau}},$$

such that

$$I=\int_{0}^{\infty}d\tau\int_{-\infty}^{\infty} \mathrm{d}^3v~3x^2y^2v_z^2\cos(uv_x) \mathrm{e^{-v_x^2(\tau y^2+1/2)-v_y^2(\tau x^2+1/2)-v_z^2(\tau x^2y^2+1/2)}}.$$

The $(v_x,v_y,v_z)$ integrals can be evaluated with the help of Mathematica. The results gives

$$\int_{-\infty}^{\infty} \mathrm{d}^3v~3x^2y^2v_z^2\cos(uv_x) \mathrm{e^{-\alpha v_x^2-\beta v_y^2-\gamma v_z^2}}=\frac{3\pi^{3/2}}{2x^2y^2}\frac{\exp\left(-\frac{u^2}{4y^2}\frac{1}{\tau+1/2y^2}\right)}{\left(\tau+1/2x^2y^2\right)^{3/2}\left(\tau+1/2x^2\right)^{1/2}\left(\tau+1/2y^2\right)^{1/2}}.$$

Thereby,

$$I= -\frac{3~\mathrm{const}}{x^2y^2~\mathrm{const}}\int_{0}^{\infty}\mathrm{d\tau}\frac{\exp\left(-\frac{u^2}{4y^2}\frac{1}{\tau+1/2y^2}\right)}{\left(\tau+1/2x^2y^2\right)^{3/2}\left(\tau+1/2x^2\right)^{1/2}\left(\tau+1/2y^2\right)^{1/2}}.$$

Now, performing the substitution $\tau=\frac{1-x^2}{2x^2y^2k^2}-\frac{1}{2x^2y^2}$ gives us

$$I=(\mathrm{const})~\frac{3x~y}{(1-x^2)^{3/2}}\int_{0}^{\sqrt{1-x^2}}\mathrm{dk}\frac{k^2\exp\left(-\frac{u^2}{2}\frac{x^2k^2}{\left(1-x^2\right)\left(1-k^2\right)}\right)}{\sqrt{1-k^2}\sqrt{1-k^2\frac{1-y^2}{1-x^2}}},$$

which is the desired integral unless of a constant. ;)