solving Improper integral I have a problem when I tried to solve the following improper integral
I think it could be solved using Residue theorem, but I don't how to crack it , and I don't know if there is a closed form for this problem
$$\int_{0}^{\infty}\frac{sin(sin(t-\theta))e^{cos(t-\theta)}}{t^{1/2}} dt$$ $\ where\ \theta >0$
I faced this problem while I'm trying to find the fractional derivative of a function using one of the definitions.
Any help would be appreciated.
 A: I'm not sure a closed form is possible, even with contour integration, but I was able to get a fairly clean result;
$$\int_{0}^{\infty}{\sin(\sin(t-\theta))\cdot e^{\cos(t-\theta)}\over t^{1\over 2}}dt=\mathfrak{Im}\left\{\int_{0}^{\infty}{e^{i(\sin(t-\theta))}\cdot e^{\cos(t-\theta)}\over t^{1\over 2}}dt\right\}$$
$$=\mathfrak{Im}\left\{\int_{0}^{\infty}{e^{e^{i(t-\theta)}}\over t^{1\over 2}}dt\right\}=\mathfrak{Im}\left\{\sum_{n=0}^{\infty}{1\over n!}\int_{0}^{\infty}{{e^{i(nt-n\theta)}}\over t^{1\over 2}}dt\right\}$$
$$=\mathfrak{Im}\left\{\sum_{n=0}^{\infty}{e^{-in\theta}\over n!}\int_{0}^{\infty}t^{-{1\over 2}}\cdot{{e^{i(nt)}}}dt\right\}$$
$$=\mathfrak{Im}\left\{\int_{0}^{\infty}{t^{-{1\over2}}}dt+\sum_{n=1}^{\infty}{e^{-in\theta}\over n!}\int_{0}^{\infty}t^{-{1\over 2}}\cdot{{e^{i(nt)}}}dt\right\}$$
The integral that we have split off from our sum diverges. However, because we are only concerned with the imaginary part, it never becomes an issue. Focusing on the term containing the infinite sum and applying the substitution $u=-int$ gives us...
$$\mathfrak{Im}\left\{\sum_{n=1}^{\infty}{e^{-in\theta}\over n!}\int_{0}^{\infty}t^{-{1\over 2}}\cdot{{e^{i(nt)}}}dt\right\}=\mathfrak{Im}\left\{i\sum_{n=1}^{\infty}{e^{-in\theta}\over n\cdot n!}\int_{0}^{\infty}\bigg({-u\over in}\bigg)^{-{1\over 2}}\cdot{{e^{-u}}}du\right\}$$
$$=\mathfrak{Im}\left\{\sqrt{i}\sum_{n=1}^{\infty}{e^{-in\theta}\over \sqrt{n}\cdot n!}\int_{0}^{\infty}{u}^{-{1\over 2}}\cdot{{e^{-u}}}du\right\}=\mathfrak{Im}\left\{\sqrt{i\pi}\sum_{n=1}^{\infty}{e^{-in\theta}\over \sqrt{n}\cdot n!}\right\}$$
$$=\mathfrak{Im}\left\{\sqrt{\pi}\left({1\over \sqrt{2}}+{i\over \sqrt{2}}\right)\sum_{n=1}^{\infty}{(\cos(n\theta)-i\sin(n\theta))\over \sqrt{n}\cdot n!}\right\}$$
Finally, some quick multiplication yields our answer;
$$=\sqrt{\pi\over2}\sum_{n=1}^{\infty}{(\cos(n\theta)-\sin(n\theta))\over \sqrt{n}\cdot n!}$$
