MM88: rectangular contour integration

[@yaozhang1687]

Obviously parameter a cannot equals or less than zero. When a=0, lim x→-∞ e^ax/1+e^x = 1; When a

[@DR_VIV]

Good answer!

[@jwkim4428]

Eve if a is negative, integrand is positive. So the integral over real axis must be positive, but your answer may be negative for negative a.

[@DR_VIV]

You are correct, of course. I should have put absolute value signs around ‘a’

[@hirakJR]

Handwriting 👌

[@DR_VIV]

Thanks!!

[@ben34256]

8:42 I don't think the denominator is correct here. It should be |1+exp(R)exp(iy)|. As |z|/|1+z| >= |z|(1/(1+|z|)) not greater than or equal.

[@DR_VIV]

You are correct, of course… the bound should still work out…. If one uses the minimum value in the denominator, which would be absolute value of R minus 1.

[@zinzhao8231]

beautiful problem!

[@Thebiggame-dc8ek]

Thank you very much. But I wonder what about taking the contour at y below the pole, for example at y=i*pi/2I, then the integral will be equal to zero, why this result can't be true? and how we know if we include more poles the result won't change like when no pole were involved?
I appreciate if you explain it to me. Much thanks.

[@DR_VIV]

Our ultimate aim is to evaluate the integral on the real line… that aim must not be lost sight of in our choice (choices) of the contour, of which as you point out, there are infinitely many. But all of them must involve the real axis or as you point out you will get zero answer by Cauchy theorem

[@igorlimarochaazevedo5536]

Why did you subtracted 1 in one of the last equations of the resolution --> 1 - e^2(2pia)?

[@DR_VIV]

That came from gathering the terms of the computation from the previous steps. It is not 1 by itself but 1 multiplied by the value of the integral we seek. So it is a common factor.

[@Dedicate25]

Please tell me the name of that pen!!

[@DR_VIV]

Pen that Video begins with is Dani trio tosca in mottled green hard rubber, fitted with a factory stub nib.

[@Dedicate25]

@@DR_VIV Thank you sir

[@kalaivani.kkannan3517]

Why take 2pi , sir

[@DR_VIV]

Any other value may also be used.

[@kalaivani.kkannan3517]

thankyou sir, i thought 2*pi*i is periodic value for e^z, @@DR_VIV