Integral between 0 and infty of {1- (1/pi) [arctan(4/x)-arctan(-4/x)] } *(1/x) dx?

2011-04-02T11:41:54Z

Hello,

I would need to calculate the integral between 0 and +infty of {1- (1/pi) [arctan(4/x)-arctan(-4/x)]}*(1/x) dx.

Could you please let me know how to do that and what's the correct answer?

Thanks loads in advance for your help;

All the best, Elise

Integrate kˆ(n-1) / prod_{i=1...n} (kˆ(2)+x_iˆ{2}) dk between 0 and infinity, with x_i constants and n>=1?

2011-04-02T01:03:39Z

[some formatting tweaked, and the question copied from the title to the main body, by YC]

Hi,

I've been struggling a lot to calculate this integral.

$$ \int_0^\infty k^{n-1} /\quad \prod_{i=1}^n (k^2+ x_i^2)\; dk $$ where $x_i$ are constants and $n\geq 1$.

I did the calculation for n=1,2,3,4, with the hope of identifying some form and then find the result by induction. But here is what I got:

n=1: I= (pi/2) * abs(x1)
n=2: I= (1/2) * 1/(x2ˆ(2)-x1ˆ(2)) * log(x2ˆ(2) / x1ˆ(2))
n=3: I= (pi/2) * [abs(x1) (x2ˆ(2)-x3ˆ(2)) +abs(x2) (x3ˆ(2)-x1ˆ(2))+ abs(x3) (x1ˆ(2)-x2ˆ(2))] / [(x2ˆ(2)-x3ˆ(2) (x3ˆ(2)-x1ˆ(2)) (x1ˆ(2)-x2ˆ(2)]
n=4: I= (1/2) * [ A1 log(x1ˆ(2)) + A2 log(x2ˆ(2)) +... A4 log(x4ˆ(2))), where Ai= xiˆ(2) / [ prod (xjˆ(2)-xiˆ(2))]

-->> This makes me think that the result depends on whether n is even or uneven; that is, we would have a form in log( ) for n even, and something in pi/2 for n uneven?

Could you please help me here? What is the correct result and how to get it?

Your help is so much appreciated, many many thanks in advance! Elise

Comment by Payze

2011-04-02T11:47:11Z

Thank you so much for your insights, much appreciated :-) Best, Elise

User payze - MathOverflow

Integral between 0 and infty of {1- (1/pi) [arctan(4/x)-arctan(-4/x)] } *(1/x) dx?

Integrate kˆ(n-1) / prod_{i=1...n} (kˆ(2)+x_iˆ{2}) dk between 0 and infinity, with x_i constants and n>=1?

Comment by Payze