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

Francisca Rodden 2022-01-04 Answered
I want to evaluate the following integral via complex analysis
x=0x=eaxcos(bx)dx, a>0
Which function/ contour should I consider ?
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Nadine Salcido
Answered 2022-01-05 Author has 34 answers
0eaxcos(bx)dx=R(0+eaxeibxdx)
(where R(z) denotes the real part of (z). Then,
0+eaxeibxdx=0+e(aib)xdx
=limM+0M0+e(aib)xdx
=limM+[1aibe(aib)x]0M
=limM+(1aibe(aib)M+1aib)
=1aib
=a+ib|aib|2
So,
0+eaxcos(bx)dx=a|aib|2
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poleglit3
Answered 2022-01-06 Author has 32 answers
Let us integrate the function eAz, where A=a2+b2 on a circular sector in the first quadrant, centered at the origin and of radius R, with angle ω which satisfies cosω=aA, and therefore sinω=bA. Let this sector be called γ.
Since our integrand is obviously holomorphic on the whole plane we get:
γdzeAz=0
Breaking it into its three pieces we obtain:
aRdxeAx+0ωdϕiReAReiϕ+R0dreiωeAreω=0
1A=1A0dr(a+ib)er(a+ib)
0dr(a+ib)ear(cosbrisinbr)=1
Now let's call Ic=0drearcosbr and Is=0drearsinbr, then:
aIciaIs+ibIc+bIs=1
and by solving:
aIc+bIs=1; aIs+bIc=0
Ic=aa2+b2; Is=ba2+b2
This method relies only on the resource of contour integration as you asked!
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star233
Answered 2022-01-11 Author has 208 answers

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(ax+ibx). Use this function to evaluate the unbounded integral then evaluate the bounded one

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