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Residue Theorem

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Need a little help with this please...
Did you try plugging it into symbolab?
Original post by 3672N

Need a little help with this please...

Consider the complex valued function f(z)=eizz+if(z) = \dfrac{e^{iz}}{z+i} and integrate around a semi-circle of radius RR centered at the origin in the lower half of the complex plane (denoted as Γ\Gamma).

Then consider the fact that, by Residue Theorem, we have:

Γf(z) dz=2πiRes(f(z),i)\displaystyle \int_{\Gamma} f(z) \ \mathrm{d}z = 2 \pi i \mathrm{Res}(f(z),-i)

Do you know how to proceed? Have you seen similar examples?
(edited 2 years ago)
Reply 3
Original post by RDKGames
Consider the complex valued function f(z)=eizz+if(z) = \dfrac{e^{iz}}{z+i} and integrate around a semi-circle of radius RR centered at the origin in the lower half of the complex plane (denoted as Γ\Gamma).

Then consider the fact that, by Residue Theorem, we have:

Γf(z) dz=2πiRes(f(z),i)\displaystyle \int_{\Gamma} f(z) \ \mathrm{d}z = 2 \pi i \mathrm{Res}(f(z),-i)

Do you know how to proceed? Have you seen similar examples?


Tried this way, but stucked where I had to equate real and imaginary parts as both the intergrals with sinx and cosx had complex denominators.
Reply 4
Original post by Anonymy0us
Did you try plugging it into symbolab?


Yes but didn't work
Reply 5
Original post by 3672N
Yes but didn't work

Can you upload what you tried? The final solution
https://www.wolframalpha.com/input/?i=integrate+sin%28x%29%2F%28x%2Bi%29+dx+
(edited 2 years ago)
Reply 6
Original post by 3672N
Tried this way, but stucked where I had to equate real and imaginary parts as both the intergrals with sinx and cosx had complex denominators.

"stucked" ???

Post your working. You should be able to equate real and imaginary parts after calculating the residue.

(You should also be able to explain why the integral round the semicircle goes to 0 as R increases - possibly you know a lemma that covers this from your lectures?)
Reply 7
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Expanding sine x simplified the intergral, Thank you all!...

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