I would like to find a "simple" series form for the following integrals:
\begin{align} I_1(x)&=\int_{t=0}^{\frac{\pi}{2}} t e^{-ix\cos{t}}\,dt\\ I_2(x)&=\int_{\phi=0}^{2\pi} \int_0^{\frac{\pi}{2}} t e^{-ix\sin{t}\sin{\phi}}\,dt \,d\phi \end{align}
Mathematica can give a series form of the integrands, but then integrating to get my final series $I_1(x)$ and $I_2(x)$ gives quite horrible coefficients involving special functions.
Is there a method likely to give a simpler form of coefficients in the series?
Many thanks!
Using $e^z=\sum_{n=0}^\infty z^n/n!$, you get $I_1(x)=\sum_{n=0}^\infty(J_n/n!)(-ix)^n$ where $J_n=\int_0^{\pi/2}t\cos^n t\,dt$.
Using integration by parts, we obtain a recurrence for $J_n$: for $n>1$ $$J_n=t\cos^{n-1}t\sin t\,\Bigg|_{0}^{\pi/2}-\int_0^{\pi/2}\big(\cos^{n-1}t-(n-1)t\cos^{n-2}t\sin t\big)\sin t\,dt\\=(n-1)(J_{n-2}-J_n)-\frac1n\implies J_n=\frac{n-1}{n}J_{n-2}-\frac{1}{n^2},$$ which is solved, using the double factorial notation (with $0!!=(-1)!!=1$), by $$\begin{aligned}J_{2n}&=\frac{(2n-1)!!}{(2n)!!}\left(\frac{\pi^2}{8}-\sum_{k=1}^{n}\frac{(2k-2)!!}{(2k-1)!!}\frac{1}{2k}\right),\\J_{2n+1}&=\frac{(2n)!!}{(2n+1)!!}\left(\frac{\pi}{2}-\sum_{k=0}^{n}\frac{(2k-1)!!}{(2k)!!}\frac{1}{2k+1}\right).\end{aligned}\qquad(n\geqslant 0)$$
The power series for $I_2(x)$ is obtained pretty much the same way.