Integral of $\int\limits_0^{2\pi } {{a^{\frac{{b\cos (x - c)}}{d}}}dx} $?

Question

I am trying to find the integral of,

$\int\limits_0^{2\pi } {{a^{\frac{{b\cos (x - c)}}{d}}}dx} $

Where, $a,b,c,d \in R$.

I am trying to find the definite integral in wolfram alpha online but it does not provide me any result. Does that mean the Integral does not exist or wrong? I could not do that. Thank you.

Answer 1

your integral: $$\mathcal{I}=\int_0^{2\pi}a^{\frac{b\cos(x-c)}{d}}dx=\int_0^{2\pi}e^{\alpha\cos(x-c)}dx=\int_c^{2\pi+c}e^{\alpha\cos x}dx$$ where $\alpha=\frac{b\ln(a)}{d}$ we can try and split this integral up now, since $\cos x$ has a period of $2\pi$ we can write this as: $$\mathcal{I}=-\int_0^cf(x)dx+\int_0^{2\pi}f(x)dx+\int_0^cf(x)dx$$ $$=\int_0^{2\pi}f(x)dx=\int_0^{2\pi}e^{\alpha\cos x}dx=2\int_0^\pi e^{\alpha\cos x}dx$$ which is a standard integral called the Bessel function of the first kind, and so your answer is: $$\mathcal{I}=2\pi I_0\left(\frac{b\ln a}{d}\right)$$

Answer 2

Hint: Assuming $a > 0$, you can use the identity: $$ I_0(x) = \frac{1}{\pi}\int_0^\pi e^{x\cos\theta}d\theta $$ Where $I_0(x)$ is a modified Bessel function of the first kind.