Does $\int _0^{\pi }e^x\sin ^n x\:\mathrm{d}x$ have a closed form? Does a closed form for $\displaystyle\int _0^{\pi }e^x\sin ^n x\:\mathrm{d}x$ exist?
I tried to evaluate this with values like $n=1,2$ with integration by parts and it seemed fine but when i tried with higher values such as $n=3,4,5$ it became more tedious and couldn't manage to evaluate.
Could you please help me find the closed form of this expression please?.
 A: $$
\int_0^\pi e^x \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^n \, dx
$$
Expand the $n$th power via the binomial theorem and apply the distributive law, then integrate term by term.
As a function of $n$ I wouldn't call that a closed form, but as a function of $x$ it is a closed form.
A: I fed Quanto's recurrence to Maple.  It says the solution is:
$$
I_n = {\frac {\pi\,{{\rm e}^{\pi/2}}n! }{{2}^{n}
\Gamma \left( n/2+1+i/2 \right) \Gamma \left( n/2+1-i/2 \right) }}.
$$
A: The formula given by @GEdgar
$$I_n=\frac{\pi\,e^{\frac \pi 2}}{2^n } \frac{  \Gamma (n+1)}{\Gamma \Big[\frac{n}{2}+\left(1-\frac{i}{2}\right)\Big]\,\Gamma \Big[\frac{n}{2}+\left(1+\frac{i}{2}\right)\Big]}$$ is perfectly correct.
Writing
$$I_{2n+1}=(e^\pi+1)\, a_n$$ the $a_n$'s form the sequence
$$\left\{\frac{1}{2},\frac{3}{10},\frac{3}{13},\frac{63}{325},\frac{2268}{13325},
   \frac{24948}{162565},\frac{149688}{1062925},\cdots\right\}$$ and writing
$$I_{2n}=(e^\pi-1)\, b_n$$ the $b_n$'s form the sequence
$$\left\{1,\frac{2}{5},\frac{24}{85},\frac{144}{629},\frac{8064}{40885},\frac{14515
   2}{825877},\frac{19160064}{119752165},\cdots\right\}$$
A: Let $I_n = \int _0^{\pi }e^x\sin ^n x dx$ and integrate  by parts twice to get the recursive equation below
$$I_n = \frac{n(n-1)}{n^2+1}I_{n-2}$$
with $I_0= e^\pi-1$ and $I_1=\frac12(1+e^\pi)$.
