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




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Hi,


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


$$ \int\_0^\infty \frac{k^{n-1}}{\prod\_{i=1}^n (k^2+ x\_i^2)}\; dk
$$
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



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?
You are correct. Use the partial fraction decomposition: <http://en.wikipedia.org/wiki/Partial_fraction> For example, if $n=4$, the decomposition is (over the rationals):


$$\begin{array}{l}
{\frac {ck}{ \left( {k}^{2}+c \right) \left( -c+a \right) \left( -c+
b \right) \left( -d+c \right) }}-\\\ {\frac {dk}{ \left( {k}^{2}+d
 \right) \left( -d+a \right) \left( -d+b \right) \left( -d+c
 \right) }}-\\\ {\frac {bk}{ \left( {k}^{2}+b \right) \left( -b+a
 \right) \left( -c+b \right) \left( -d+b \right) }}+\\\ {\frac {ak}{
 \left( {k}^{2}+a \right) \left( -b+a \right) \left( -c+a \right)
 \left( -d+a \right) }}\end{array}
$$
where $a=x\_1^2,b=x\_2^2,...$. I guess you can get the pattern. For odd $n$ there is no $k$ in the numerator. This is the cause of the difference you noticed. 


