I want to evaluate the following integral via complex analysis ∫x=0x=∞e−axcos⁡(bx)dx, a&gt;0 Which function/ contour should I consider ?

Question

I want to evaluate the following integral via complex analysis  ∫x=0x=∞e−axcos⁡(bx)dx, a&amp;gt;0  Which function/ contour should I consider ?

Nadine Salcido · Accepted Answer

∫0∞e−axcos⁡(bx)dx=R(∫0+∞e−axeibxdx)  (where R(z) denotes the real part of (z). Then,  ∫0+∞e−axeibxdx=∫0+∞e−(a−ib)xdx  =limM→+∞∫0M∫0+∞e−(a−ib)xdx  =limM→+∞[−1a−ibe−(a−ib)x]0M  =limM→+∞(−1a−ibe−(a−ib)M+1a−ib)  =1a−ib  =a+ib|a−ib|2  So,  ∫0+∞e−axcos⁡(bx)dx=a|a−ib|2

poleglit3 · Accepted Answer

Let us integrate the function e−Az, 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 cos⁡ω=aA, and therefore sin⁡ω=bA. Let this sector be called γ.Since our integrand is obviously holomorphic on the whole plane we get:∮γdze−Az=0Breaking it into its three pieces we obtain:∫aRdxe−Ax+∫0ωdϕiRe−AReiϕ+∫R0dreiωe−Areω=01A=1A∫0∞dr(a+ib)e−r(a+ib)∫0∞dr(a+ib)e−ar(cos⁡br−isin⁡br)=1Now let&#039;s call Ic=∫0∞dre−arcos⁡br and Is=∫0∞dre−arsin⁡br, then:aIc−iaIs+ibIc+bIs=1and by solving:aIc+bIs=1; −aIs+bIc=0Ic=aa2+b2; Is=ba2+b2This method relies only on the resource of contour integration as you asked!

star233 · Accepted Answer

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

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

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