# How can I calculate an integral, perhaps using residues?

Could you advise me on how to calculate the following integral, perhaps by using Residue?

Integrate[(1.4142*Sqrt[s*(0.1*s + Sqrt[4 + 0.01*s^2])])/
(s^3 Sqrt[4 + 0.01*s^2] (-78.9568 + s*(0.1*s + Sqrt[4 + 0.01*s^2])))
(1 - s*Sqrt[4 + 0.01*s^2]) Cos[(Pi + t) s]*
Sin[0.707*x*Sqrt[s*(0.1*s + Sqrt[4 + 0.01*s^2])]],
{s, 0., Infinity},
PrincipalValue -> True,
Assumptions -> t + Pi > 0]

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Please try to be more specific and provide an example integral which you'd like to compute (preferably in Mathematica code as well), and also show us what you have tried. – Leonid Shifrin Dec 24 '12 at 17:56
It's more of a math question, because if you now how residues work from complex variable theory you know how to use Residue to calculate improper integrals. – swish Dec 24 '12 at 18:05
@swish I can see here a meaningful question on how Mathematica can be used to automate some parts of these computations, but the question still should IMO be more specific. – Leonid Shifrin Dec 24 '12 at 18:08
Well then, I should describe the problem that I get. I need to plot the following improper integral Plot3D[Integrate[( 1.4142* Sqrt[s*(0.1*s + Sqrt[ 4 + 0.01*s^2])])/(s^3 Sqrt[ 4 + 0.01*s^2] (-78.9568 + s*(0.1*s + Sqrt[ 4 + 0.01*s^2]))) (1 - s*Sqrt[ 4 + 0.01*s^2]) Cos[(Pi + t)s] * Sin[0.707*x*Sqrt[s* (0.1*s + Sqrt[ 4 + 0.01*s^2])]], {s, 0., Infinity}, PrincipalValue -> True, Assumptions -> t + Pi > 0],{t,-Pi,Pi},{x,0,1}]. But have not any result. So, I thought compute this integral using Residues and then try plot. – user5166 Dec 24 '12 at 18:21
Something like this maybe? Total[2 \[Pi] I Residue[f[x], {x,x /. #}] & /@ Solve[Denominator[f[x]] == 0 && Im[x] > 0, x] – swish Dec 24 '12 at 18:27