Help solving an improper integral with recursion of some sort

Question

So i have this integral:

$$\int_{0}^{\infty} e^{-x}\sin^nxdx$$

So I am not allowed to use the reduction formulae as a fact, without giving a proof, well maybe for $sin^n x$ only, but okay.

My first move is:

$$\int_{0}^{\infty} e^{-x}\sin^nxdx= (\sin^nx(-e^{-x}) \Big|_0^{\infty} - n\int_{0}^{\infty} \cos x\sin^{n-1}x(- e ^{-x})dx$$

But if I continue doing this, I would get the same thing as in the start, except with $\cos^nx$ instead of $\sin^nx$. So perhaps I did the per partes part wrong, or choose the incorrect path. Also in noticed that $e^{-x}$ on the "upper bound", will go to 0, and since integration doesn't affect $e^x$ as much is it safe to assume it will just go to 0? Or am I thinking about it inccorectly?

If anyone could help me get through this integral I would deeply apreciate it.

Thank you in advance.

Answer 1

Let

$$I_n = \int_0^\infty e^{-x}\sin ^n x\ dx$$

Consider just the second, integration term in your question,

$$\begin{align*} J &= -n\int_0^\infty \cos x\sin^{n-1}x\left(-e^{-x}\right)dx\\ &= -n\int_0^\infty \cos x \sin^{n-1}x \ de^{-x}\\ &= \left.-ne^{-x}\cos x\sin^{n-1}x\right|_0^{\infty} +n\int_0^\infty e^{-x}d\left(\cos x\sin^{n-1}x\right)\\ &= {\left.-ne^{-x}\cos x\sin^{n-1}x\right|_0^{\infty} + {n\int_0^\infty e^{-x}\left[\cos x (n-1)\sin^{n-2}x\cos x + \sin^{n-1}x(-\sin x)\right]dx}}\\ &= {\left.-ne^{-x}\cos x\sin^{n-1}x\right|_0^{\infty} + {n\int_0^\infty e^{-x}\left[(n-1)\sin^{n-2}x(1-\sin^2 x) + \sin^{n-1}x(-\sin x)\right]dx}}\\ &= \left.-ne^{-x}\cos x\sin^{n-1}x\right|_0^{\infty} + n(n-1)\int_0^\infty e^{-x}\sin^{n-2}x\ dx - n^2\int_0^\infty e^{-x}\sin^n x\ dx\\ &= \left.-ne^{-x}\cos x\sin^{n-1}x\right|_0^{\infty} + n(n-1)I_{n-2} - n^2 I_n \end{align*}$$

Combining with your result,

$$\begin{align*} I_n &= \left[-e^{-x}\sin ^n x-ne^{-x}\cos x\sin^{n-1}x\right]_0^{\infty} + n(n-1)I_{n-2} - n^2 I_n\\ &= \frac{\left[-e^{-x}\sin ^n x-ne^{-x}\cos x\sin^{n-1}x\right]_0^{\infty} + n(n-1)I_{n-2}}{1+n^2} \end{align*}$$

Compare with reduction formulae table on Wikipedia.

Answer 2

The integral has a closed form, depending on the parity of $n$. When $n=2m$ we have the following: $$\int_0^\infty e^{-x}\sin^{2m}x\ dx =\frac{(-1)^m}{4^m}\sum_{k=0}^{2m}{{2m} \choose k}\frac{(-1)^k}{1+4(m-k)^2}$$ I am still working on the odd case.