Help needed with difficult integral $\int_{-\pi}^{\pi}\cos(a \cos(x)) e^{b \cos(x'-x)} dx $

Question

I have quite a tricky integral to be solved. I tried already several things, but I couldn't find the solution yet. To know that it's definitely not analytically solvable would help as well for not spending more time than necessary on it.

The integral that needs to be solved is:

$\int_{-\pi}^{\pi}\cos(a \cos(x)) \exp(b \cos(x'-x)) \mathrm{d}x $

Ideas I had so far were:

The parts of the integral could at least be integrated from $-\pi$ to $\pi$.

Where $\int_{-\pi}^{\pi}\cos(a \cos(x)) \mathrm{d}x = 2\pi \mathrm{J}_0(|a|)$ when $a $ is real. With $\mathrm{J}_0(|a|)$ being a Bessel function of the first kind.

$\int_{-\pi}^{\pi}\exp(b \cos(x'-x)) = 2\pi \mathrm{I}_0(b)$ with $\mathrm{I}_0(b)$ being a modified Bessel function of the first kind.

My first idea was to do a integration by parts, but this does not work as there is no indefinite integral for both parts, or?

Second idea was to transform the first cosinus also into exponentials and try to solve them together, but I still have some problems in bringing $\cos(x)$ and $\cos(x' -x)$ together.

Third option is to do some expansions and try to use some orthogonality relations. I have seen something similar, when someone was using the Jacobi-Anger expansion to solve

$\int_{-\pi}^{\pi}\exp(\mathrm{i}m\phi)) \exp(b \cos(\phi' -\phi)) \mathrm{d}\phi = 2\pi\mathrm{I}_m(b) \exp(\mathrm{i}m\phi)$

from the expansion they got $\exp(b \cos \theta) = \sum_{n = -\infty}^{\infty} \mathrm{I}_n(b) \exp(\mathrm{i}n\theta) $ Then they wrote something about that $\exp(\mathrm{i}m\phi),m \in \mathbb{Z} $ being the eigenvalues of the Fredholm integral operator and that everything vanishes except for $n=m$ .

Does anyone understand what's going on there and if I could use that for my problem?

Thanks for your help!

EDIT after Jack's answer:

Using the cosine sum formula $\cos(x' -x) = \cos(x')\cos(x)+\sin(x')\sin(x)$ is something I tried already, but I had problems with it. It gave me the following equation:

$\int_{-\pi}^{\pi}\cos(a \cos(x)) \exp(b' \cos(x))\exp(b'' \sin(x)) \mathrm{d}x $ with $b'=b\cos(x')$ and $b'' = b\sin(x')$.

I also get, that I can write $\cos(t)$ as $\Re(\exp(\mathrm{i}t))$, but now there's the point where I always get stuck. Every part of the integral can be solved on it's own, but I can not separate them. Usually I would try integration by parts, but as I said before there's the problem, that none of the terms has an indefinite integral. Probably it can be solved with this translation thing, but I don't know anything about it.

Could you just make your answer a little bit more explicit within this regard (to show how one can write my integral in terms of $C(a,b)$ and $S(a,b)$) or give me a link to somewhere where this translation method is explained?

Thanks again!

Answer 1

Using a translation we have: $$I(a,b)=\int_{-\pi}^{\pi}\cos(a \cos(x))\,e^{b\cos(x'-x)}dx = \int_{-\pi}^{\pi}\cos(a \cos(x-x'))\,e^{b\cos(x)}dx =$$ $$=\int_{-\pi}^{\pi}\cos(a\cos(x')\cos(x)+a\sin(x')\sin(x)))\,e^{b\cos(x)}dx=$$ $$=\Re\left(\int_{-\pi}^{\pi}\exp\left((b+iA_1)\cos x+iA_2\sin x\right)dx\right),$$ where $A_1=a\cos(x')$ and $A_2=a\sin(x')$. If $x'$ belong to $\frac{\pi}{2}\mathbb{Z}$, we only have to develop expressions for: $$C(a,b)=\Re\left(\int_{-\pi}^{\pi}\exp\left((b+ia)\cos x\right)dx\right),\quad S(a,b)=\Re\left(\int_{-\pi}^{\pi}\exp\left(b\cos x+ia \sin x\right)dx\right).$$ By considering the Taylor series of the exponential function we have: $$C(a,b)=\Re\left(\sum_{j=0}^{+\infty}\int_{-\pi}^{\pi}\frac{(b+ia)^{2j}\cos^{2j}x}{(2j)!}dx\right)=\Re\left(2\pi\sum_{j=0}^{+\infty}\frac{(b+ia)^{2j}\binom{2j}{j}}{4^j(2j)!}\right),$$ $$C(a,b)=2\pi\sum_{j=0}^{+\infty}\frac{(a^2+b^2)^j \cos\left(2j\arctan(a/b)\right)}{4^j (j!)^2}=2\pi\sum_{j=0}^{+\infty}\frac{(a^2+b^2)^j\cdot T_{2j}\left(\frac{b}{\sqrt{a^2+b^2}}\right)}{4^j (j!)^2}$$ $$ C(a,b)=2\pi\cdot\Re\left(I_0(b+ia)\right).$$ In a similar way: $$S(a,b)=2\pi\sum_{j=0}^{+\infty}\frac{(b^2-a^2)^j}{4^j(j!)^2}=2\pi\cdot I_0\left(\sqrt{b^2-a^2}\right).$$ In the general case: $$I(a,b)=2\pi\cdot\Re\left(\sum_{j=0}^{+\infty}\frac{(b^2-a^2+2iab\cos(x'))^{j}}{4^j(j!)^2}\right)=2\pi\cdot\Re\left(I_0\left(\sqrt{b^2-a^2+2iab\cos(x')}\right)\right).$$

Answer 2

The solution Jack gave me was giving me the right direction, but was not explicit enough to see right away (at least for me) what was going on and some parts are probably not necessery. So I'll try to answer my question again myself in a more detailed way to see, if I got everything right... So if you spot any mistakes please let me know.

Again the integral to be solved was:

$I(a,b) = \int_{-\pi}^{\pi}\cos(a \cos(x)) \mathrm{e}^{b \cos(x'-x)} \mathrm{d}x $

With $\cos(x) =\Re(\mathrm{e}^{\mathrm{i}x})$ I get

$I(a,b) = \int_{-\pi}^{\pi} \Re\left( \mathrm{e}^{ia \cos(x)}\right) \mathrm{e}^{b \cos(x - x')} \mathrm{d}x$

as $\mathrm{e}^{b \cos(x - x')}$ is real I can write

$I(a,b) = \Re\left( \int_{-\pi}^{\pi} \mathrm{e}^{ia \cos(x)+ b \cos(x - x')} \mathrm{d}x\right)$.

Now I'll use $\mathrm{e}^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$ and get

$I(a,b) = \Re\left( \int_{-\pi}^{\pi} \sum_{n=0}^{\infty} \frac{(ia \cos(x)+ b \cos(x - x'))^n}{n!} \mathrm{d}x\right)$.

Now the next step I'm not 100% sure I'm allowed to do it like this: I found the following identity-formula on wikipedia

(http://en.wikipedia.org/wiki/Trigonometric_identity#Linear_combinations):

$a\sin x+b\sin(x+\alpha)= c \sin(x+\beta)\,$

where

$c = \sqrt{a^2 + b^2 + 2ab\cos \alpha},\,$

and

$ \beta = \arctan \left(\frac{b\sin \alpha}{a + b\cos \alpha}\right) + \left\{\begin{array}{cl} 0 &\text{if } a + b\cos \alpha \ge 0 \\ \pi & \mbox{if } a + b\cos \alpha < 0 \end{array}\right. $

Does this also work if a and b are complex numbers? Could anyone give me a hint of a citeable textbook with the identity for complex a and b? For the a and b real I found it also in Bronstein chapter 2.5.2.1.4. Probably the distinction of cases has to be adapted, but this will not influence on my calculation, I hope, so I'll continue for now.

$I(a,b) = \Re\left( \int_{-\pi}^{\pi} \sum_{n=0}^{\infty} \frac{(c^n \sin^n(x+\beta)}{n!} \mathrm{d}x\right)$.

with $c= \sqrt{b^2 - a^2 + 2iab\cos x'}$.

To solve

$I(a,b) = \Re\left( \sum_{n=0}^{\infty} \frac{ c^n \int_{-\pi}^{\pi}(\sin^n(x+\beta) \mathrm{d}x}{n!}\right)$.

the next step is to calculate the integrals

$i(a,b) = \int_{-\pi}^{\pi}( \sin^n(x+\beta) \mathrm{d}x$ for all $n$.

Now a distinction of cases for $n$ has to be made.

If $n$ is even we get together with the following power-reduction formula

(http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formula) $\sin^n\theta = \frac{1}{2^n} \binom{n}{\frac{n}{2}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} \binom{n}{k} \cos{((n-2k)\theta)}$

$i(a,b) = \int_{-\pi}^{\pi} \frac{1}{2^n} \binom{n}{\frac{n}{2}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} \binom{n}{k} \cos{((2i)(x+\beta))} \mathrm{d}x$

where $m = n-k $.

$i(a,b) = \int_{-\pi}^{\pi} \frac{1}{2^n} \binom{n}{\frac{n}{2}} \mathrm{d}x + \int_{-\pi}^{\pi} \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} \binom{n}{k} \cos{(2mx+\beta')} \mathrm{d}x$ with $\beta'=2m\beta$ which is $i(a,b) = \frac{2\pi}{2^n} \binom{n}{\frac{n}{2}} $ as $\int_{-\pi}^{\pi} \cos{(2mx+\beta')} \mathrm{d}x =0$ as $\sin(2\pi m +\phi)=\sin(\phi) $, $ m \in \mathbb{N} $

and if $k$ is odd $i(a,b)=0$ therefore.

$I(a,b) = 2\pi\Re\left( \sum_{n=0}^{\infty} \frac{ \left(\frac{c}{2}\right)^{2n} \binom{2n}{n} }{(2n)!} \right) = 2\pi\Re\left( \sum_{n=0}^{\infty} \frac{ \left(\frac{c}{2}\right)^{2n} }{n!n!} \right) = 2\pi\Re\left( \mathrm{I}_0(c)\right)$ using the following equations for the modified Bessel function of the first kind $\mathrm{I}_\nu=\sum_{r=0}^{\infty}\frac{\left(\frac{x}{2}\right)^{2r+\nu}}{\Gamma(r+\nu+1)r!}$ and the Gamma function $\Gamma(n) = (n-1)! $ when $ n \in \mathbb{N} $.

So finally we got the result $I(a,b) = 2\pi\Re\left(\mathrm{I}_0(\sqrt{b^2 - a^2 + 2iab\cos x'})\right)$