## Bit of a complicated integral, need some help

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### Bit of a complicated integral, need some help

For personal use I am trying to solve the following integral:

\int_0^1 \frac {\mathrm{d}t} {\prod\limits_{i=0}^n [(x_B t - x_i)^2 + y_i^2]}

After some deliberation, I decided that the easiest way to go about this is to split up the fraction using the method of partial fractions. This yields the following (assuming I got my syntax right):

1 = \sum\limits_{i=0}^{n} [ A_i \prod\limits_{j \in n - \{i\}} [ (x_b t - x_j)^2 + y_i^2 ]]

And then if we substitute \frac { x_i - y_i } { x_B } \sqrt{-1} for t for each i, we'll get a system of linear equations, which then need to be solved for all of the Ai...

... and that's where it goes a bit beyond my grasp. Is there any way I can proceed, or have I gotten in too deep?
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Aedl Foxe

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### Re: Bit of a complicated integral, need some help

First, I think your system of equations is not quite right; because the denominators are quadratic in t the numerators are generally linear in t, so you want (Ai+Bit) there.

But there's a quicker way (than setting up the linear system of equations) to find the partial-fraction expansion, in the special case where all of the denominators are linear polynomials,
\frac{P(t)}{Q(t)} = \sum_m \frac{A_m}{t-t_m}\;.

Multiply the equation through by a factor (t-tk) and evaluate it at t=tk. All of the partial fractions on the right side give zero, except for the kth one. So the right-hand side is just Ak, and you immediately have a formula for the Ak.

In your case you can factor each of the quadratic factors into a complex-conjugate pair to give the desired linear factors,
\frac{A_m+B_mt}{(x_Bt+x_m)^2+y_m^2} = \frac{C_m}{x_Bt+x_m+iy_m} + \frac{D_m}{x_Bt+x_m-iy_m}\;.

Do the above to get Cm and Dm (you should find that Dm=Cm* the complex conjugate), then combine these complex-conjugate partial-fraction pairs to get Am and Bm (which should both be real). Then you can integrate this to give a combination of a logarithm and an arctangent.

eta oin shrdlu

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