Solving an integral with a parameter : $ I(a) = \int_{0}^{\infty} \frac{a^{\cos(x)}\sin(\sin(x)\ln {a})}{x}dx$

Question

The integral $$ I(a) := \int_{0}^{\infty} \frac{a^{\cos(x)}\sin(\sin(x)\ln {a})}{x}dx$$ I got this integral from a friend. I tried to solve this, but was unable to find the solution. Even WolframAlpha is unable to provide an answer.
My attempts:

$1$) Using the complex definition of $ \sin(x)$, the integral becomes:

$$ I(a) = \Re{\frac{1}{2i}\int_{0}^{\infty}\frac{e^{\cos(x)\ln{a}}(e^{i \sin(x)\ln {a}}-e^{-i \sin(x) \ \ln {a}})}{x}}dx$$

which becomes (I am skipping a few steps):

$$I(a) = \Re{\frac{1}{2i}\int_{0}^{\infty}\frac{a^{e^{ix}}-a^{e^{-ix}}}{x}}dx$$

Then I differentiated under the integral

$$I'(a) = \Re{\frac{1}{2i}\int_{0}^{\infty}\frac{\partial}{\partial a}\frac{a^{e^{ix}}-a^{e^{-ix}}}{x}}$$

$$I'(a) = \Re{\frac{1}{2}\int_{0}^{\infty}\frac{((ae)^{ix}+(ae)^{-ix})}{x}}$$ I am not sure what to do after this. The Gamma function seems to be popping up in the second part of the integral.

Is there any other way to compute this integral more cleanly, or just in general, how to solve this integral?

Answer 1

As @RandomVariable said it is a generalization of this integral so I will write an answer to my own ques.
We have this integral.

$$I(a) = \int_{0}^{\infty}\frac{e^{ln {a}\cos(x)}\sin(\sin{x}\ln {a})}{x}dx$$ $$I(a) = \Im{\int_{0}^{\infty}\frac{e^{\ln {a}(\cos{x}+i\sin{x})}}{x}}dx$$ $$I(a) = \int_{0}^{\infty}\frac{\Im{e^{\ln {a}(e^{ix})}}}{x}dx$$ $$I(a) = \int_{0}^{\infty}\Im{\sum_{n>0}\frac{(\ln{a})^{n}e^{inx}}{n!x}}dx$$ $$I(a) = \sum_{n>0}\frac{(\ln{a})^{n}}{n!}\int_{0}^{\infty}\frac{\sin{nx}}{x}dx$$

We know that $\int_{0}^{\infty}\frac{\sin{nx}}{x} = \frac{\pi}{2}$ therefore

$$I(a) = \frac{\pi}{2}\sum_{n>0}\frac{(\ln {a})^{n}}{n!}$$ $$I(a) = \frac{\pi}{2}(a-1)$$