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- Sep 15th 2013, 09:05 PMthevinhSome challenge integrals to think about.

- Sep 16th 2013, 05:44 PMtopsquarkRe: Some challenge integrals to think about.
Do all of these have exact solutions? That is to say exact without needing 12 terms involving logarithms and arctangents. I had to use the full Mathematica package to do the first two...Wolfram|Alpha just wan't up to it.

-Dan - Sep 16th 2013, 06:02 PMthevinhRe: Some challenge integrals to think about.
Yes, all of them have exact solution and can be integrated using man power. No need to use Mathematica or Wolfram. Our "super-computer" is much more capable and sophisticated than those softwares. In addition, no need to utilize advance concepts such as complex integration etc.... My answer to #1 is .

- Sep 16th 2013, 06:08 PMthevinhRe: Some challenge integrals to think about.
- Sep 17th 2013, 09:41 AMSworDRe: Some challenge integrals to think about.
The first 3 can all be done with brute force, simply by using partial fractions, breaking the function into terms of the form , then integrating term by term.

I used a quicker method for #1 though, and I got . Then I looked up that the answer was Pi/3, and interestingly my answer does break down to pi/3 with the use of some identities. - Sep 17th 2013, 10:10 AMFelixFelicis28Re: Some challenge integrals to think about.
- Sep 17th 2013, 10:19 AMSworDRe: Some challenge integrals to think about.
- Sep 17th 2013, 11:22 AMFelixFelicis28Re: Some challenge integrals to think about.
- Sep 17th 2013, 11:35 AMSworDRe: Some challenge integrals to think about.
I was just finishing that one but you beat me to it :o

- Sep 17th 2013, 12:09 PMFelixFelicis28Re: Some challenge integrals to think about.
- Sep 17th 2013, 02:25 PMSworDRe: Some challenge integrals to think about.
- Sep 17th 2013, 04:53 PMFelixFelicis28Re: Some challenge integrals to think about.
- Sep 17th 2013, 08:35 PMthevinhRe: Some challenge integrals to think about.
- Sep 17th 2013, 09:22 PMSworDRe: Some challenge integrals to think about.
Here's how I did it:

http://latex.codecogs.com/png.latex?...^4+1}{x^6+1}dx

(notice the change in starting index)

Let

The contour integral of this function around a circle of increasing radius approaches 0, so the sum of the residues must be 0. The residues at any integers are exactly the terms in the series, so the sum of the series is the sum of the residues at z=-5/12 and z=-11/12, times negative 1. This is a common approach in complex analysis for summing series.

http://latex.codecogs.com/png.latex?...%20)\right%20)

Now I must admit, compared to the solutions above this method is rather unusual. But it is interesting because it arrives at a result which is impressed in a different form, yet equal to .