How to calculate $\int_{0}^{\infty}e^{-ax}\sin^nx~\mathrm dx$ I have no idea how to evaluate 
$$\int_{0}^{\infty}e^{-ax}\sin^nx~\mathrm dx$$
In Table of Integrals,Series and Products  3.895，I found a formula about,but I don't know how to prove it.
 A: I do not know what sort of formula they gave in the Table you mention, but here is the way I done this one a good while back. It was in some old papers I knew I had somewhere :)
EDIT:  I noticed marty cohen mentions this in a comment. 
Use parts:
$$u=\sin^{n}(x), \;\ dv=e^{-ax}dx, \;\ v=\frac{-1}{p}e^{-ax}, \;\ du=n\sin^{n-1}(x)\cos(x)dx$$
$$I_{n}=\frac{-1}{a}e^{-ax}\sin^{n}(x)|_{0}^{\infty}+\frac{n}{a}\int_{0}^{\infty}\sin^{n-1}(x)\cos(x)e^{-ax}dx$$
$$I_{n}=\frac{n}{a}\int_{0}^{\infty}e^{-ax}\sin^{n-1}(x)\cos(x)dx$$
Parts again:
$$\begin{align} I_{n}=\frac{n}{a}\left[\frac{-1}{a}e^{-ax}\sin^{n-1}(x)\cos(x)|_{0}^{\infty}+\int_{0}^{\infty}\frac{e^{-ax}}{a}\cdot \left((n-1)\sin^{n-2}(x)\cos^{2}(x)-\sin^{n}(x)\right)dx\right]\end{align} $$
with $\begin{align} \cos^{2}(x)=1-\sin^{2}(x) \end{align}$:
$$\begin{align}I_{n}=\frac{n(n-1)}{a^{2}}\int_{0}^{\infty}e^{-ax}\sin^{n-2}(x)dx-\frac{n^{2}}{a^{2}}\int_{0}^{\infty}e^{-ax}\sin^{n}(x)dx\end{align}$$
So, we can now write:
$$I_{n}=\frac{n(n-1)}{a^{2}}I_{n-2}-\frac{n^{2}}{a^{2}}I_{n}$$
solve for $I_{n}$:
$$I_{n}\left(1+\frac{n^{2}}{a^{2}}\right)=\frac{n(n-1)}{a^{2}}I_{n-2}$$
$$I_{n}=\frac{n(n-1)}{n^{2}+a^{2}}I_{n-2}$$
Now, if n is even, the recursion results in:
$$I_{n \;\ \text{even}}=\frac{n(n-1)}{n^{2}+a^{2}}\cdot \frac{(n-2)(n-3)}{(n-2)^{2}+a^{2}}\cdot\cdot\cdot \frac{2\cdot 1}{2^{2}+a^{2}}\cdot \underbrace{\frac{1}{a}}_{\text{I_{0}}}$$
When n is odd:
$$I_{n \;\ \text{odd}}=\frac{n(n-1)}{n^{2}+a^{2}}\cdot \frac{(n-2)(n-3)}{(n-2)^{2}+a^{2}}\cdot \cdot \cdot \frac{3\cdot 2}{3^{2}+a^{2}}\cdot \underbrace{\frac{1}{a^{2}+1}}_{\text{I_{1}}}$$
Look back and we can quickly notice that $$I_{0}=\frac{1}{a}, \;\ I_{1}=\frac{1}{1+a^{2}}$$
Thus, we ultimately have for even n:
$$\displaystyle I_{n \;\ \text{even}}=\frac{n!}{\displaystyle \prod_{k=1}^{n/2}\left(4k^{2}+a^{2}\right)}$$
$$=\frac{\pi a\cdot csch(\frac{\pi a}{2})n!}{2^{n+1}\Gamma(\frac{n}{2}-\frac{ai}{2}+1)\Gamma(\frac{n}{2}+\frac{ai}{2}+1)}$$ 
For odd n:
$$\displaystyle I_{n \;\ \text{odd}}=\frac{n!}{\displaystyle \prod_{k=0}^{\frac{n-1}{2}}\left((2k+1)^{2}+a^{2}\right)}$$
$$=\frac{\pi sech(\frac{\pi a}{2})\cdot n!}{2^{n+1}\Gamma(\frac{n}{2}-\frac{ai}{2}+1)\Gamma(\frac{n}{2}+\frac{ai}{2}+1)}$$
Make sure I have no typos or oversights in all of this.
A: $$\sin^nx=\frac1{(2i)^n}\left(e^{ix}-e^{-ix}\right)^n$$
Expand the brackets using Pascal's Triangle.  Write the results involving $\cos kx$ and $\sin kx$, for $k$ between $0$ and $n$.  Then do the Laplace transform on each one.
