Evaluate $\int t^2 e^{-2i\pi nt}\,dt$ I need to get
$$\int t^2 e^{-2i\pi nt}\,dt$$
I'm thinking to use integration by parts, but $\int e^{-2i\pi nt}\,dt$ is tripping me up. Can anybody help? Thanks!
 A: Integration by parts will lead you to the answer
Use

$$\int e^{ax}\, dx=\frac{e^{ax}}{a}+C$$

and You'll have
$$\int e^{-2i\pi nt} \,dt= -\frac{e^{-2i\pi nt}}{2i\pi n}+C$$
Final result would be

$$\int t^2e^{2i\pi nt}\,dt=\frac{e^{2i\pi nt}(-2i\pi^2 n^2 t^2+2\pi nt+ i)}{4\pi^3 n^3}+C$$

A: You can set $2i\pi n=\alpha$, then 
$$
\begin{eqnarray}
\int t^2 e^{-\alpha t}\mathrm dt&=&\int \frac{d^2 e^{-\alpha t}}{d\alpha^2} \mathrm dt=\frac{d^2}{d\alpha^2}\int e^{-\alpha t}\mathrm dt=-\frac{d^2}{d\alpha^2}\left(\frac{e^{-\alpha  t}}{\alpha }\right)=\\ &=&-\frac{(i-(1+i) \pi  n t) (1+(1+i) \pi  n t)}{4 \pi ^3 n^3}e^{-2 i \pi  n t}.
\end{eqnarray}
$$
A: Hint:
$2in\pi t=z \implies dt=\dfrac{dz}{2in\pi}$
$\therefore\displaystyle\int t^2e^{-2in\pi t}dt=-\left(\dfrac{1}{8in^3\pi^3}\right)\displaystyle\int z^2 e^{z}\ dz$
A: Observe that
\begin{align}
\frac{d}{dt} t^2 e^{-2\pi i nt}&=2t e^{-2\pi i nt}-2\pi i n t^2 e^{-2\pi i nt},\\
\frac{d}{dt} t e^{-2\pi i nt}&= e^{-2\pi i nt}-2\pi i n t e^{-2\pi i nt},\\
\frac{d}{dt} e^{-2\pi i nt}&= -2\pi i n e^{-2\pi i nt}.
\end{align}
Thus
\begin{align}
t^2 e^{-2\pi i nt} &= \frac{2}{2\pi i n}t e^{-2\pi i nt} -  \frac{d}{dt} \frac{t^2}{2\pi i n} e^{-2\pi i nt} \\
&= \frac{2}{(2\pi i n)^2} e^{-2\pi i nt} -  \frac{d}{dt} \left(\frac{t^2}{2\pi i n} e^{-2\pi i nt}+\frac{2t}{(2\pi i n)^2}te^{-2\pi i nt}\right)\\
&=-\frac{d}{dt}\left(\frac{t^2}{2\pi i n}+\frac{2t}{(2\pi i n)^2}+\frac{2}{(2\pi i n)^3}\right) e^{-2\pi i nt}.
\end{align}
