how to evaluate $\int_{0}^{+\infty} e^{-at}(sin(t))^{n} dt$

Question

How to evaluate the following definite integral : $$\int_{0}^{+\infty} e^{-at}\left(\sin(t)\right)^{n}\,dt$$

Answer 1

I give you some hints: Let $$ I_n=\int_0^{+\infty}e^{-at}\sin^n t\,dt=\mathcal L(\sin^n t)(a), $$ i.e. $I_n$ is the Laplace transform of $\sin^n(t)$. Using the fact that $$ D^2(\sin^nt)=n(n-1)\sin^{n-2}t-n^2\sin^n t. $$ we find that, for $n>1$, (using the rule for differentiation and Laplace transforms) $$ a^2I_n=n(n-1)I_{n-2}-n^2I_n, $$ or $$ I_n=\frac{n(n-1)}{n^2+a^2}I_{n-2}. $$ Here, we have also used the fact that $\sin^nt$ is zero for $t=0$ and that its derivative is zero for $t=0$.

Now, as is easily calculated $$ I_0=\frac{1}{a},\quad\text{and}\quad I_1=\frac{1}{1+a^2}. $$ From this it follows that $$ I_n=\frac{n!}{a(2^2+a^2)(4^2+a^2)\cdots(n^2+a^2)},\quad \text{if $n$ is even} $$ and $$ I_n=\frac{n!}{(1^2+a^2)(3^2+a^2)\cdots(n^2+a^2)},\quad\text{if $n$ is odd.} $$

Answer 2

Use De Moivre's formula to compute the Fourier sine/cosine series of $\sin(t)^n$, then exploit: $$ \int_{0}^{+\infty}e^{-at}\sin(bt)\,dt = \frac{b}{a^2+b^2},\qquad \int_{0}^{+\infty}e^{-at}\cos(bt)\,dt = \frac{a}{a^2+b^2}.$$