integration of exp(cos(x-a)) dx

Question

I would like to compute either of the following integrals: $$\int e^{\cos(x-a)} \, dx$$ or $$\int_{-\pi}^{\pi} e^{\cos(x-a)} \, dx$$

In both cases, $a$ is a constant.

MATLAB doesn't seem to be helpful with either.

Answer 1

The integrand is periodic, with period $2\pi$, so the second one is independent of $a$. In fact $$ \int_{-\pi}^\pi e^{\cos(x-a)}\,dx = 2\pi I_0(1) \approx 7.9549 $$ where $I_0$ is a certain Bessel function.

In the integral form for $I_\alpha(x)$ given on that page, put $x=1$ and $\alpha=0$.

Answer 2

For $\int e^{\cos(x-a)}~dx$ ,

$\int e^{\cos(x-a)}~dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{\cos^{2n}(x-a)}{(2n)!}dx+\int\sum\limits_{n=0}^\infty\dfrac{\cos^{2n+1}(x-a)}{(2n+1)!}dx$

$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\cos^{2n}(x-a)}{(2n)!}\right)dx+\int\sum\limits_{n=0}^\infty\dfrac{\cos^{2n+1}(x-a)}{(2n+1)!}dx$

For $n$ is any natural number,

$\int\cos^{2n}(x-a)~dx=\dfrac{(2n)!x}{4^n(n!)^2}+\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin(x-a)\cos^{2k-1}(x-a)}{4^{n-k+1}(n!)^2(2k-1)!}+C$

This result can be done by successive integration by parts.

For $n$ is any non-negative integer,

$\int\cos^{2n+1}(x-a)~dx$

$=\int\cos^{2n}(x-a)~d(\sin(x-a))$

$=\int(1-\sin^2(x-a))^n~d(\sin(x-a))$

$=\int\sum\limits_{k=0}^nC_k^n(-1)^k\sin^{2k}(x-a)~d(\sin(x-a))$

$=\sum\limits_{k=0}^n\dfrac{(-1)^kn!\sin^{2k+1}(x-a)}{k!(n-k)!(2k+1)}+C$

$\therefore\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\cos^{2n}(x-a)}{(2n)!}\right)dx+\int\sum\limits_{n=0}^\infty\dfrac{\cos^{2n+1}(x-a)}{(2n+1)!}dx$

$=x+\sum\limits_{n=1}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin(x-a)\cos^{2k-1}(x-a)}{4^{n-k+1}(n!)^2(2k-1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\sin^{2k+1}(x-a)}{(2n+1)!k!(n-k)!(2k+1)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin(x-a)\cos^{2k-1}(x-a)}{4^{n-k+1}(n!)^2(2k-1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\sin^{2k+1}(x-a)}{(2n+1)!k!(n-k)!(2k+1)}+C$