Difficult Fourier Transform involving $e^{-cx}\sin bx$

Question

I have to compute the Fourier transform of the function $\displaystyle f(x) = \begin{cases} e^{-cx}\sin bx & \text{if}\, x > 0 \\ 0 & \text{if}\, x<0 \end{cases}$

where $c$ and $b$ are positive constants. I also should check that $f \in L^{2}(-\infty, \infty)$.

For both parts of this, I get to a point and then get stuck and discouraged.

I tried to directly calculate the Fourier transform using the formula $\displaystyle F(\alpha) = \frac{1}{2\pi}\int_{-\infty}^{\infty} f(x) e^{-i \alpha x}dx$:

$\displaystyle \begin{align} \frac{1}{2\pi} \int_{0}^{\infty} e^{-cx}\sin bx e^{-i \alpha x}dx = \frac{1}{2\pi}\int_{0}^{\infty}\sin bx e^{-(c+i\alpha)x}dx \end{align}$

From here, I tried converting $\displaystyle \sin(bx) = \frac{e^{ibx} - e^{-ibx}}{2i}$ to make the integral solely in terms of exponentials:

$\displaystyle \begin{align} \frac{1}{2\pi}\int_{0}^{\infty}\sin bx e^{-(c+i\alpha)x}dx = \frac{1}{2\pi}\int_{0}^{\infty}\frac{(e^{ibx}-e^{-ibx})}{2i} \cdot e^{-(c+i\alpha)x}dx \\=\frac{1}{4\pi i}\int_{0}^{\infty}\left(e^{ibx}e^{-(c+i\alpha x)} - e^{-ibx}e^{-(c+i\alpha)x}\right) dx \\ = \frac{1}{4 \pi i}\int_{0}^{\infty} \left(e^{ibx-cx-i\alpha x}-e^{-ibx-cx-i\alpha x}\right)dx \\ = \frac{1}{4 \pi i}\int_{0}^{\infty}\left(e^{-cx+(bx-\alpha x)i}-e^{-cx-(bz+az)i}\right)dx \\ = \frac{1}{4 \pi i}\int_{0}^{\infty}\left( e^{(-c+(b-a)i)x}-e^{(-c-(b+a)i)x}\right)dx \\ = \frac{1}{4 \pi i}\int_{0}^{\infty} e^{(-c+(b-a)i)x}dx - \frac{1}{4 \pi i}\int_{0}^{\infty}e^{(-c-(b+a)i)x} dx \\ = \frac{1}{4 \pi i}\lim_{d \to \infty}\int_{0}^{d} e^{(-c+(b-a)i)x}dx - \frac{1}{4 \pi i}\lim_{d \to \infty}\int_{0}^{d}e^{(-c-(b+a)i)x} dx\end{align}$

After evaluating each of these integrals and taking the limits as $d \to \infty$, I get that $\displaystyle F(\alpha) = \frac{1}{2\pi} \frac{b}{(c-(b-a)i)(c+(b+a)i)}$,

a far cry from the $\mathbf{\displaystyle F(\alpha) = \frac{b}{a^{2}-(c-b)^{2}}}$ given as the answer in the back of the book.

I am an absolute mess. Where did I go wrong??

Originally, I thought about using one of the shifting properties of Fourier transforms that my book mentions: $\mathcal{F}[e^{icx}f(x)]=F(\alpha - c)$, but the $e^{-cx}$ factor in this problem doesn't have an imaginary part, so that didn't help me.

I'm so lost, and I need to learn how to do these. This is not homework, I'm just trying to teach myself how to do Fourier transforms. I haven't seen a lot of worked out examples, so giving me a full solution wouldn't necessarily be a bad thing, but you can tell me whatever you want so long as it helps me master Fourier transforms.

Thank you.

Answer 1

I get the same as your result. Maple agrees as well.

If you want to use shifting instead, start with the Fourier transform of (with $\theta(x)$ being the Heaviside step function) $$ \mathscr{F}\big( \exp(-cx)\theta(x) \big) = \frac{1}{2\pi}\frac{1}{c+i\alpha}. $$ Hence $$ \mathscr{F}\big( e^{\pm ibx} \exp(-cx)\theta(x) \big) = \frac{1}{2\pi}\frac{1}{c+i(\alpha\mp b)}. $$ and you end up with the same result after subtracting and dividing by $2i$.

Answer 2

Hopefully that is not the answer in the back of the book as it should be a function of $\alpha$, not a constant.

If $\operatorname{re} a >0$, then $\int_0^\infty e^{-ax} dx = {1 \over a}$.

Then we have $\hat{f}(\alpha) = {1 \over 2\pi} {1 \over 2i} \int_0^\infty (e^{-x[c+i(\alpha -b)]}- e^{-x[c+i(\alpha +b)]} ) dx = {1 \over 2\pi} {1 \over 2i} ( {1 \over c+i(\alpha -b)} - {1 \over c+i(\alpha +b)} ) = { 1\over 2 \pi} {b \over (c+ i \alpha)^2+b^2}$.