Finding a Closed Form for $\int_0^\infty A\sin(\frac{2\pi}{T}x) \exp(-bx)dx$ I'm struggling in finding a closed form for, 

$$ \int_0^\infty Ae^{-bx}\sin\left(\frac{2\pi}{T}x\right) \,dx $$

My Attempt: Let $H = \int Ae^{-bx}\sin\left(\frac{2\pi}{T}x\right)\,dx$. First I integrate by parts, splitting $f=e^{-bx}$ and $dg=\sin\left(\frac{2\pi}{T}x\right)\,dx$. Then another IBP using $u=e^{-bx}$ and $dv = \cos\left(\frac{2\pi}{T}x\right)\,dx$ and simplify,
\begin{align}
\frac{2\pi}{A\cdot T} H &= -e^{-bx}\cos\left(\frac{2\pi}{T}x\right) - \frac 1 b \int \cos\left(\frac{2\pi}{T}x\right)e^{-bx}\,dx \\
\frac{2\pi}{A\cdot T} H &= -e^{-bx}\cos\left(\frac{2\pi}{T}x\right) - \frac 1 b \left[ e^{-bx}\frac{T}{2\pi}\sin\left(\frac{2\pi}{T}x\right) + \frac{T}{2\pi b}\int e^{-bx}\sin\left(\frac{2\pi}{T}x\right)\,dx \right] \\
\left(\frac{2\pi}{A\cdot T} - \frac{T}{2\pi b^2}\right)H &= e^{-bx}\cos\left(\frac{2\pi}{T}x\right) - \frac{T}{2\pi b} e^{-bx} \sin\left(\frac{2\pi}{T}x\right) \\
H &= -e^{-bx} \left(\cos\left(\frac{2\pi}{T}x\right) + \frac{T}{2\pi b}\sin\left(\frac{2\pi}{T}x\right) \right) \cdot \left( \frac{A\cdot T 2\pi b^2}{(2\pi b)^2 - A\cdot T^2}\right) + c\\
\end{align}
First off, is this horrid looking expression correct? If so, can it be simplified further?
Now I come to evaluate $H|_0^\infty$. $H$ appears to be defined at $x=0$, so I really just need to find $\lim_{x\to \infty^+} H$. Here is where I'm stuck. The denominator grows exponentially for $b\ge 1$, and $T$ and $A$ don't seem to have any effect on the limit. How should I evaluate this limit?

This is the Laplace Transform of $\sin \left(\dfrac{2 \pi  x}{T}\right)$
Its value is $\dfrac{2 \pi  A T}{b^2 T^2+4 \pi ^2}$
I just looked on the Laplace transform table
