Finding a Closed Form for $\int_0^\infty A\sin(\frac{2\pi}{T}x) \exp(-bx)dx$

Question

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?

Answer 1

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

Answer 2

Even without Laplace transform, consider the two antiderivatives $$I=\int \sin(ax) e^{-bx}\,dx \qquad \text{and} \qquad J=\int \cos(ax) e^{-bx}\,dx$$ $$J+iI=\int e^{iax} e^{-bx}\,dx=\int e^{-(b-ia)x}\,dx=-\frac{e^{ -(b-ia)x}}{b-ia}\tag 1$$ $$J-iI=\int e^{-iax} e^{-bx}\,dx=\int e^{-(b+ia)x}\,dx=-\frac{e^{-(b+ia)x}}{b+ia}\tag 2$$ $$(J+iI)-(J-iI)=2iI=-\frac{e^{ -(b-ia)x}}{b-ia}+\frac{e^{-(b+ia)x}}{b+ia}\tag 3$$ Manipulate the complex to get $$2iI=-\frac{2 i e^{-b x} (b \sin (a x)+a \cos (a x))}{a^2+b^2}$$ that is to say $$I=-\frac{ b \sin (a x)+a \cos (a x)}{a^2+b^2}e^{-b x}$$ So, $$K=\int_0^\infty \sin(ax) e^{-bx}\,dx=\frac{a}{a^2+b^2}$$ provided that $|\Im(a)|<\Re(b)$.

Replace $a$ by $\frac {2\pi}T$ to get the result.