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


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


closed as off-topic by Ricardo Andrade, Chris Godsil, Ryan Budney, Andrey Rekalo, Dmitri Pavlov Jan 20 '14 at 16:19

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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.

  • $\begingroup$ Thank you so much for your insights, much appreciated :-) Best, Elise $\endgroup$ – Payze Apr 2 '11 at 11:47

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