# I want to evaluate the following integral via complex analysis \int_{x=0}^{x=\infty}e^{-ax}\cos(bx)dx,\

I want to evaluate the following integral via complex analysis

Which function/ contour should I consider ?
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${\int }_{0}^{\mathrm{\infty }}{e}^{-ax}\mathrm{cos}\left(bx\right)dx=R\left({\int }_{0}^{+\mathrm{\infty }}{e}^{-ax}{e}^{ibx}dx\right)$
(where $R\left(z\right)$ denotes the real part of (z). Then,
${\int }_{0}^{+\mathrm{\infty }}{e}^{-ax}{e}^{ibx}dx={\int }_{0}^{+\mathrm{\infty }}{e}^{-\left(a-ib\right)x}dx$
$=\underset{M\to +\mathrm{\infty }}{lim}{\int }_{0}^{M}{\int }_{0}^{+\mathrm{\infty }}{e}^{-\left(a-ib\right)x}dx$
$=\underset{M\to +\mathrm{\infty }}{lim}{\left[-\frac{1}{a-ib}{e}^{-\left(a-ib\right)x}\right]}_{0}^{M}$
$=\underset{M\to +\mathrm{\infty }}{lim}\left(-\frac{1}{a-ib}{e}^{-\left(a-ib\right)M}+\frac{1}{a-ib}\right)$
$=\frac{1}{a-ib}$
$=\frac{a+ib}{{|a-ib|}^{2}}$
So,
${\int }_{0}^{+\mathrm{\infty }}{e}^{-ax}\mathrm{cos}\left(bx\right)dx=\frac{a}{{|a-ib|}^{2}}$
###### Not exactly what you’re looking for?
poleglit3
Let us integrate the function ${e}^{-Az}$, where $A=\sqrt{{a}^{2}+{b}^{2}}$ on a circular sector in the first quadrant, centered at the origin and of radius R, with angle $\omega$ which satisfies $\mathrm{cos}\omega =\frac{a}{A}$, and therefore $\mathrm{sin}\omega =\frac{b}{A}$. Let this sector be called $\gamma$.
Since our integrand is obviously holomorphic on the whole plane we get:
${\oint }_{\gamma }dz{e}^{-Az}=0$
Breaking it into its three pieces we obtain:
${\int }_{a}^{R}dx{e}^{-Ax}+{\int }_{0}^{\omega }d\varphi iR{e}^{-AR{e}^{i\varphi }}+{\int }_{R}^{0}dr{e}^{i\omega }{e}^{-Ar{e}^{\omega }}=0$
$\frac{1}{A}=\frac{1}{A}{\int }_{0}^{\mathrm{\infty }}dr\left(a+ib\right){e}^{-r\left(a+ib\right)}$
${\int }_{0}^{\mathrm{\infty }}dr\left(a+ib\right){e}^{-ar}\left(\mathrm{cos}br-i\mathrm{sin}br\right)=1$
Now let's call ${I}_{c}={\int }_{0}^{\mathrm{\infty }}dr{e}^{-ar}\mathrm{cos}br$ and ${I}_{s}={\int }_{0}^{\mathrm{\infty }}dr{e}^{-ar}\mathrm{sin}br$, then:
$a{I}_{c}-ia{I}_{s}+ib{I}_{c}+b{I}_{s}=1$
and by solving:

This method relies only on the resource of contour integration as you asked!
###### Not exactly what you’re looking for?
star233

First,you may use the definition of Lapace transform to get it or integration by parts twice.
For complex numbers your integrand is the real part of $exp\left(-ax+ibx\right)$. Use this function to evaluate the unbounded integral then evaluate the bounded one