Complex Analysis: Integral of log(x)/(x+1)^2

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Today, we integrate log(x)/(x+1)^2 from 0 to infinity using the keyhole contour.

qncubed3

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This channel is awesome af, please continue and don't stop, we'll always support you.

Thor-ykcr

It's me, when you obtained I=0 : 🤔
And ...
It's me, at the end of video: 🤓🤓
Great, as always (especially the second part, when you proved those integral are equal by Zero) ...
Thank you so much dear *QN³*

wuyqrbt

For a simpler method, substitute u=1/x, move around some things and replace u with x and you'll see that the integral is equal to its negative

buddychumpalfriendhomiebud

Saving me complex final grade. Idk why we are doing so many hard contours for a one semester class. My school is jamming complex 1 and 2 into a semester lmaooo

austinvera

It may be a good example to illustrate contour integration. However if you set J(a) = ∫Log[a*x]/(1+x)^2, (same limits), then
J ' [a]= 1/a *∫1/(1+x)^2 (0 to inf ) = 1/a . Thus J[a]= Log[a], but for a=1 we find J[1] = 0.

renesperb

Perfect explanation! … “in particular” the limits part of the video !

saeida.alghamdi

How draw graph complex function in complex plane

ayushtharollno

This is not very convincing. The limits in the end for big gamma are not the same as for little gamma - since r remains in the denominator and goes to infinity in the second case. The argument for big gamma to yield zero is that log r “swared” grows slower than r - and that needs to be shown. The reference to the calculations before seems confusing to me.

matthias

Great!

Can we try to evaluate ∫(0→∞)(ln(x) /(1-x^2) dx using the keyhole contour? Best regards.

George-ijgm

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