How to integrate this integral from Fourier transform?

Question

The question is from Fourier transform, but I don't know how to integrate this:

$$ \int_0^\infty \dfrac{e^{jt}+e^{-jt}}{2} e^{\large-t(1+2\pi\alpha j)} dt. $$

Answer 1

I assume $j$ is an imaginary number, $j=\sqrt{-1}$. \begin{align} \int_0^\infty \dfrac{e^{jt}+e^{-jt}}{2} e^{\large-t(1+2\pi\alpha j)} dt&=\frac{1}{2}\int_0^\infty e^{\large-t(1+2\pi\alpha j-j)}\ dt+\frac{1}{2}\int_0^\infty e^{\large-t(1+2\pi\alpha j+j)}\ dt \end{align} Let $u=t(1+2\pi\alpha j-j)$ then $du=(1+2\pi\alpha j-j)\,dt$ and $v=t(1+2\pi\alpha j+j)$ then $dv=(1+2\pi\alpha j+j)\,dt$. \begin{align} &\frac{1}{2(1+2\pi\alpha j-j)}\int_0^\infty e^{\large-u}\ du+\frac{1}{2(1+2\pi\alpha j+j)}\int_0^\infty e^{\large-v}\ dv\\ =&\frac{1}{2(1+2\pi\alpha j-j)}+\frac{1}{2(1+2\pi\alpha j+j)}\\ =&\frac{1+2\pi\alpha j+1+2\pi\alpha j-j}{2(1+2\pi\alpha j-j)(1+2\pi\alpha j+j)}\\ =&\frac{1+2\pi\alpha j}{(1+2\pi\alpha j)^2-j^2}\\ =&\frac{1+2\pi\alpha j}{1+4\pi\alpha j-4\pi^2\alpha^2+1}\\ =&\frac{1+2\pi\alpha j}{2(1-2\pi^2\alpha^2+2\pi\alpha j)} \end{align} The final result agrees with Wolfram Alpha.

Answer 2

Notice that $\dfrac{e^{jt}+e^{-jt}}{2}=\cos t$ and let $1+2\pi \alpha j=k$, then the integral you are dealing with is: $$I=\int_0^{\infty} \cos t \,e^{-kt}\,dt=\frac{k}{k^2+1}$$ The above is easy to show using integration by parts. Substitute back $k=1+2\pi \alpha j$ to get: $$I=\frac{1+2\pi \alpha j}{(1+2\pi \alpha j)^2+1}=\frac{1+2\pi \alpha j}{1+4\pi \alpha j-4\pi^2\alpha^2+1}=\frac{1+2\pi \alpha j}{2(1-2\pi^2 \alpha^2 +2\pi\alpha j)}$$ $\blacksquare$