# Difficult trigonometric integral

This question was also asked here and here.

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.

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

• Why the downvote? – Dinesh Shankar Aug 22 '18 at 0:58
• I guess the downvote could be because the title is rather uninformative, and sounds much like the titles of many low-quality questions that appear on math.stackexchange.com from students wanting homework solutions. A more informative title could be something like “How to derive Mathematica’s solution for this integral in 3-d polar co-ordinates involving nested trig functions?” – Peter LeFanu Lumsdaine Aug 31 '18 at 12:41
• @PeterLeFanuLumsdaine Thanks for the suggestion. But Mathematica failed to do the integral $(1)$. So, there's no " Mathematica’s solution". – Dinesh Shankar Aug 31 '18 at 12:49
• Dinesh: Fair enough, I had misunderstood what you wrote about what Mathematica gave. But my main point was just that it’s good to make the title as specific and informative as possible :-) – Peter LeFanu Lumsdaine Aug 31 '18 at 12:58

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 $(v_x,v_y,v_z)$ 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. ;)

• What's the answer? – David Roberts Aug 30 '18 at 6:31
• @DavidRoberts Follow these steps. in the end you will get the integral $I_G$. Please, let me know if this is clear. – Dinesh Shankar Aug 30 '18 at 9:41
• I don't own Mathematica, nor do I care enough to go and figure out the actual integral. But this answer will be most useful for future readers if you give us the output. – David Roberts Aug 30 '18 at 10:18
• Thank you. I will improve my answer. – Dinesh Shankar Aug 30 '18 at 10:23

Here is an outline for a possible approach for the first integral: expand $e^{a\cos^2x}=\sum_ka^2\frac{\cos^{2k}x}{k!}$. So, focus on $J_k:=\int_0^{\pi/2}\frac{\cos^{2k}x}{b^2\cos^2x+c^2\sin^2}dx$. Notice that $I_k:=\int_0^{2\pi}\frac{\cos^{2k}x}{b^2\cos^2x+c^2\sin^2}dx=4J_k$. This allows for contour integration over the circle $\vert z\vert=1$, using complex analysis, by letting $z=e^{ix}$: $$I_k=\frac1{4^{n-1}i}\int_{\vert z\vert=1}\frac{(z^2+1)^{2n}}{(\alpha z^2+\beta)(\beta z^2+\alpha)z^{2n-1}}\,dz$$ where $\alpha=b+c$ and $\beta=b-c$.

I would leave the details to you.

• Thank you for your time. This approach was also proposed by Jack in Mse. I'm not familiar with contour integration, but Mathematica can do the integral $J_k$. It gives: $$\frac{1}{2} \pi \sec (\pi k) \left(\frac{\left(\frac{c^2}{b^2}\right)^{k+\frac{1}{2}} \left(1-\frac{c^2}{b^2}\right)^{-k}}{c^2}-\frac{\sqrt{\pi } \, _2\tilde{F}_1\left(\frac{1}{2},1;\frac{3}{2}-k;\frac{c^2}{b^2}\right)}{b^2 \Gamma (k)}\right)$$ It seems not useful to me :( – Dinesh Shankar Aug 22 '18 at 9:14