Evaluate $\int t^2 e^{-2i\pi nt}\,dt$

Question

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!

Answer 1

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

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

Answer 2

Hint:

$2in\pi t=z \implies dt=\dfrac{dz}{2in\pi}$

$\therefore\displaystyle\int t^2e^{-2in\pi t}dt=-\left(\dfrac{1}{8in^3\pi^3}\right)\displaystyle\int z^2 e^{z}\ dz$

Answer 3

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}

Answer 4

You can set $2i\pi n=\alpha$, then $$ \begin{eqnarray} \int t^2 e^{-\alpha t}\mathrm dt&=&\int \frac{d^2 e^{-\alpha t}}{d\alpha^2} \mathrm dt=\frac{d^2}{d\alpha^2}\int e^{-\alpha t}\mathrm dt=-\frac{d^2}{d\alpha^2}\left(\frac{e^{-\alpha t}}{\alpha }\right)=\\ &=&-\frac{(i-(1+i) \pi n t) (1+(1+i) \pi n t)}{4 \pi ^3 n^3}e^{-2 i \pi n t}. \end{eqnarray} $$