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? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T09:23:36Z http://mathoverflow.net/feeds/question/60334 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60334/integrate-kn-1-prod-i1-n-k2x-i2-dk-between-0-and-infinity-wit 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? Payze 2011-04-02T01:03:39Z 2011-04-02T03:16:59Z <p>[some formatting tweaked, and the question copied from the title to the main body, by YC]</p> <hr> <p>Hi,</p> <p>I've been struggling a lot to calculate this integral.</p> <p><code>$$\int_0^\infty k^{n-1} /\quad \prod_{i=1}^n (k^2+ x_i^2)\; dk$$</code> where $x_i$ are constants and $n\geq 1$.</p> <p>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:</p> <ul> <li><p>n=1: I= (pi/2) * abs(x1) </p></li> <li><p>n=2: I= (1/2) * 1/(x2ˆ(2)-x1ˆ(2)) * log(x2ˆ(2) / x1ˆ(2))</p></li> <li><p>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)]</p></li> <li><p>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))]</p></li> </ul> <p>-->> 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? </p> <p>Could you please help me here? What is the correct result and how to get it? </p> <p>Your help is so much appreciated, many many thanks in advance! Elise</p> http://mathoverflow.net/questions/60334/integrate-kn-1-prod-i1-n-k2x-i2-dk-between-0-and-infinity-wit/60339#60339 Answer by Mark Sapir for 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? Mark Sapir 2011-04-02T02:16:32Z 2011-04-02T03:16:59Z <p>You are correct. Use the partial fraction decomposition: <a href="http://en.wikipedia.org/wiki/Partial_fraction" rel="nofollow">http://en.wikipedia.org/wiki/Partial_fraction</a> For example, if $n=4$, the decomposition is (over the rationals):</p> <p>$$\begin{array}{l} {\frac {ck}{ \left( {k}^{2}+c \right) \left( -c+a \right) \left( -c+ b \right) \left( -d+c \right) }}-\\ {\frac {dk}{ \left( {k}^{2}+d \right) \left( -d+a \right) \left( -d+b \right) \left( -d+c \right) }}-\\ {\frac {bk}{ \left( {k}^{2}+b \right) \left( -b+a \right) \left( -c+b \right) \left( -d+b \right) }}+\\ {\frac {ak}{ \left( {k}^{2}+a \right) \left( -b+a \right) \left( -c+a \right) \left( -d+a \right) }}\end{array}$$ 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. </p>