Which function/ contour should I consider ?

Francisca Rodden
2022-01-04
Answered

I want to evaluate the following integral via complex analysis

${\int}_{x=0}^{x=\mathrm{\infty}}{e}^{-ax}\mathrm{cos}\left(bx\right)dx,\text{}a0$

Which function/ contour should I consider ?

Which function/ contour should I consider ?

You can still ask an expert for help

Nadine Salcido

Answered 2022-01-05
Author has **34** answers

(where

So,

poleglit3

Answered 2022-01-06
Author has **32** answers

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:

${\int}_{a}^{R}dx{e}^{-Ax}+{\int}_{0}^{\omega}d\varphi iR{e}^{-AR{e}^{i\varphi}}+{\int}_{R}^{0}dr{e}^{i\omega}{e}^{-Ar{e}^{\omega}}=0$

$\frac{1}{A}=\frac{1}{A}{\int}_{0}^{\mathrm{\infty}}dr(a+ib){e}^{-r(a+ib)}$

${\int}_{0}^{\mathrm{\infty}}dr(a+ib){e}^{-ar}(\mathrm{cos}br-i\mathrm{sin}br)=1$

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:

$a{I}_{c}+b{I}_{s}=1;\text{}-a{I}_{s}+b{I}_{c}=0$

$I}_{c}=\frac{a}{{a}^{2}+{b}^{2}};\text{}{I}_{s}=\frac{b}{{a}^{2}+{b}^{2}$

This method relies only on the resource of contour integration as you asked!

Since our integrand is obviously holomorphic on the whole plane we get:

Breaking it into its three pieces we obtain:

Now let's call

and by solving:

This method relies only on the resource of contour integration as you asked!

star233

Answered 2022-01-11
Author has **208** answers

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