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An integral that illustrates the beauty of contour integration methods – Residue Theorem and friends

http://residuetheorem.com/2014/01/27/an-integral-that-illustrates-the-beauty-of-contour-integration-methods

Residue Theorem and friends. A compendium of difficult calculus problems, some of which lean on the residue theorem. An integral that illustrates the beauty of contour integration methods. I attack many of the integrals in this blog using contour integration methods, some of which are obvious and some of which take a little more imagination. Here is one that illustrates the beauty an power of such methods as much as any other integral here. Is to derive the integral representation. For $ a le pi/2$.

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Inverse Laplace Transform with a coinciding pole and branch point – Residue Theorem and friends

http://residuetheorem.com/2015/03/07/inverse-laplace-transform-with-a-coinciding-pole-and-branch-point

Residue Theorem and friends. A compendium of difficult calculus problems, some of which lean on the residue theorem. Inverse Laplace Transform with a coinciding pole and branch point. Recently, the following Laplace transform. Was asked to be inverted:. F(s) = s {-a-1} e {-s a} $. Because of the branch point, the way to attack a problem like this is via Cauchy’s theorem on a properly distorted Bromwich contour. Here, we want our contour to avoid the branch point at $z=0$. This, we consider. C 5$ is the c...

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Uncategorized – Residue Theorem and friends

http://residuetheorem.com/category/uncategorized

Residue Theorem and friends. A compendium of difficult calculus problems, some of which lean on the residue theorem. A method of evaluating a double integral that nobody taught you in school. May 4, 2016 – 3:19 am. The problem is to evaluate $ prod {n=2} { infty} left (1 frac{(-1) {n-1} {a n} right ) $ where $ a n = n! Sum {k=1} {n-1} frac{(-1) {k-1} {k! This is one of those cases where trying out a bunch of numbers really isn’t going to help all that much. The $a n$ look kind of like $n! I stumbled onto...

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Algebraically difficult integral – Residue Theorem and friends

http://residuetheorem.com/2014/02/18/algebraically-difficult-integral

Residue Theorem and friends. A compendium of difficult calculus problems, some of which lean on the residue theorem. The integral to evaluate is. Int {2 pi} 0 d theta dfrac{ cos 2 theta}{ 2e {i theta}-z 2} , qquad mbox {when} , z neq 2.$. Write $ zeta=e {i theta}$ as usual to get, for the integral,. I frac14 oint { zeta =1} frac{d zeta}{ zeta} frac{( zeta zeta {-1}) 2}{ 2 zeta-z 2} = -i frac14 oint { zeta =1} frac{d zeta}{ zeta 2} frac{( zeta 2 1) 2}{(2 zeta-z)(2 - bar{z} zeta)}$. Int 0 {2 pi} d theta fr...

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An integral involving a quadratic phase – Residue Theorem and friends

http://residuetheorem.com/2014/02/08/an-integral-involving-a-quadratic-phase

Residue Theorem and friends. A compendium of difficult calculus problems, some of which lean on the residue theorem. An integral involving a quadratic phase. This one was first posted on the site Integrals and Series. And was brought to my attention on M.SE. By Cody. My solution involves a contour integration, although the approach is far from trivial. Yet again, the solution boils down to finding a contour and a function to integrate over the contour. The problem involves evaluating. Oint C dz ;f(z) $.

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Adventures in integration, University Edition – Residue Theorem and friends

http://residuetheorem.com/2015/06/06/adventures-in-integration-university-edition

Residue Theorem and friends. A compendium of difficult calculus problems, some of which lean on the residue theorem. Adventures in integration, University Edition. I got a request from an old friend whom I didn’t even know existed. He is the son of my grand-advisor, if that makes any sense. He teaches, among others, a course in Real Analysis at a university in Australia. I know I must like this guy because he says stuff like this:. After all, do we teach students how to use slide rules these days? Being ...

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Inverse Laplace Transforms and Delta Functions – Residue Theorem and friends

http://residuetheorem.com/2015/06/06/inverse-laplace-transforms-and-delta-functions

Residue Theorem and friends. A compendium of difficult calculus problems, some of which lean on the residue theorem. Inverse Laplace Transforms and Delta Functions. Problem: find the inverse Laplace transform of. Well, this one should be easy and one wonders why we are even bothering. Just split $F$ up as follows:. F(s) = frac1{s-1} 1$. The ILT of $1/(s-1)$ is simply $e t$, and the ILT of $1$ is $ delta(t)$. Done. Why indeed is the ILT of $1$ equal to $ delta(t)$? The clue is that $s/(s-1)$ doesn’t...

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An odd-looking integral of a square root of trig functions – Residue Theorem and friends

http://residuetheorem.com/2014/01/19/an-odd-looking-integral-of-a-square-root-of-trig-functions

Residue Theorem and friends. A compendium of difficult calculus problems, some of which lean on the residue theorem. An odd-looking integral of a square root of trig functions. The integral to evaluate. Int 0 { pi/3} big( sqrt{3} cos x- sin x) sin x big) {1/2} cos x ,dx $. You can start by noting that. Sqrt{3} cos{x} – sin{x} = 2 sin{ left ( frac{ pi}{3}-x right)}$. Which means that the integral is. Sqrt{2} int 0 { pi/3} dx , cos{x} sqrt{ sin{x} sin{ left ( frac{ pi}{3}-x right)} = sqrt{2} I$. 7 I = sqrt...

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Integral of function with deceptive triple pole – Residue Theorem and friends

http://residuetheorem.com/2014/01/15/integral-of-function-with-deceptive-triple-pole

Residue Theorem and friends. A compendium of difficult calculus problems, some of which lean on the residue theorem. Integral of function with deceptive triple pole. Came from Hermite’s integral representation of the Hurwitz zeta function. The integral is not the most difficult that I have ever evaluated, but is interesting from a pedagogical point of view. The problem is to evaluate. Int 0 { infty} dx frac{x}{(e x-1) (x 2 4 pi 2) 2}$. Oint C dz frac{z log{z} {(e z-1)(z 2 4 pi 2) 2}$. When $k lt 0$, howe...