Floating in the complex world

Calculus Level 3

∫ − ∞ ∞ x sin ⁡ x ( x 2 + a 2 ) ( x 2 + b 2 ) d x = ? \large \int_{-\infty}^{\infty} \dfrac{x \sin x}{(x^{2}+a^{2})(x^{2}+b^{2})} dx = ?

π ( e − a − e − b ) b 2 − a 2 \frac {\pi(e^{-a}-e^{-b})}{b^2-a^2} ( b 2 − a 2 ) ( e − a − e − b ) π \frac {(b^2-a^2)(e^{-a}-e^{-b})}\pi π \pi e π \frac e \pi

Charan Sankar by Charan Sankar , 18, United Kingdom

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2 solutions

Chew-Seong Cheong
Feb 19, 2018

I = ∫ − ∞ ∞ x sin ⁡ x ( x 2 + a 2 ) ( x 2 + b 2 ) d x Using the infinite upper semicircle = ∫ C − i z e i z ( z 2 + a 2 ) ( z 2 + b 2 ) d z of complex plain as contour = ∫ C − i z e i z b 2 − a 2 ( 1 z 2 + a 2 − 1 z 2 + b 2 ) d z = − i b 2 − a 2 ∫ C ( z e i z ( z + i a ) ( z − i a ) − z e i z ( z + i b ) ( z − i b ) ) d z Only z = i a , z = i b lie within contour. = 2 π b 2 − a 2 ( i a e − a 2 i a − i b e − b 2 i b ) By residue theorem = π ( e − a − e − b ) b 2 − a 2 \begin{aligned} I & = \int_{-\infty}^{\infty} \frac {x\sin x}{(x^2+a^2)(x^2+b^2)} dx & \small \color{#3D99F6} \text{Using the infinite upper semicircle} \\ & = \int_C \frac {-ize^{iz}}{(z^2+a^2)(z^2+b^2)} dz & \small \color{#3D99F6} \text{of complex plain as contour} \\ & = \int_C \frac {-ize^{iz}}{b^2-a^2}\left(\frac 1{z^2+a^2}-\frac 1{z^2+b^2}\right) dz \\ & = \frac {-i}{b^2-a^2} \int_C \left(\frac {ze^{iz}}{(z+ia)(z-ia)}-\frac {ze^{iz}}{(z+ib)(z-ib)}\right) dz & \small \color{#3D99F6} \text{Only }z=ia, z=ib \text{ lie within contour.} \\ & = \frac {2\pi}{b^2-a^2} \left(\frac {iae^{-a}}{2ia}-\frac {ibe^{-b}}{2ib}\right) & \small \color{#3D99F6} \text{By residue theorem} \\ & = \boxed{\dfrac {\pi(e^{-a}-e^{-b})}{b^2-a^2}} \end{aligned}

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Nice solution!!!

Charan Sankar - 3 years, 3 months ago

Sir can u tell me from where to get these types of question and how to understand it easily as I find it a little difficult to understand

A Former Brilliant Member - 3 years, 3 months ago

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I learned it long ago and am relearning it here. All my recent learning is with Brilliant.org and the internet. You should look for books or ask you teachers. I refer to this link . I am also trying to get more reference.

Chew-Seong Cheong - 3 years, 3 months ago
Charan Sankar
Feb 18, 2018

res(f,ia) = e^-a /2(b^2-a^2) res(f,ib) = -e^-b /2(b^2-a^2) Therefore, the integral is equal to pi/(b^2-a^2) * (e^-a-e^-b)

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