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muppet
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Hi everyone, this isn't really a homework question- this integral has come up during project work- but this seems like a sensible place to ask it nontheless.
Compute
I(t)=\int^{\infty}_{0} dx \frac{x^5}{(t+x^2)(1+(t+x^2)^4)}
where t is real and positive.
If f(z)=h(z)/g(z), and g has a simple pole at z=z_0, then
Res(f,z_0) = \frac{h(z_0)}{g'(z_0)}
\oint_ Cf(z)dz=2\pi i \sum_{j}Res(f,z_j)+\pi i \sum_{k} Res(f,b_k)
where C is a contour enclosing poles at z=z_j that "infinitesmally semi-circles around" simple poles at z=b_k
As my integral only runs over the positive real axis, I can't integrate around the "usual" semi-circle of infinite radius in the upper half-plane. However, as the integrand is odd in x (or, changing language to the conventions of complex analysis, in z), I think I can take the real part of an integral around a quarter-circle of infinite radius along the positive real axis and down the negative imaginary axis. This gives me a simple pole on the boundary at z=i\sqrt{t} and a simple pole at z=\sqrt{e^{i \pi/4}-t}. Plugging these in and taking the real part gives me
I(t)=Re \left[ \frac{\pi i}{4}e^{-i \pi /2}(e^{i \pi /4}-t)^{2} \right] =\frac{\pi}{4}(t^2-t\sqrt{2})
This is unfortunately wrong- mathematica gives me a decreasing function of t, but in a horribly complicated form, whilst I happen to know that a much simpler form exists. Any help would be greatly appreciated.
Homework Statement
Compute
I(t)=\int^{\infty}_{0} dx \frac{x^5}{(t+x^2)(1+(t+x^2)^4)}
where t is real and positive.
Homework Equations
If f(z)=h(z)/g(z), and g has a simple pole at z=z_0, then
Res(f,z_0) = \frac{h(z_0)}{g'(z_0)}
\oint_ Cf(z)dz=2\pi i \sum_{j}Res(f,z_j)+\pi i \sum_{k} Res(f,b_k)
where C is a contour enclosing poles at z=z_j that "infinitesmally semi-circles around" simple poles at z=b_k
The Attempt at a Solution
As my integral only runs over the positive real axis, I can't integrate around the "usual" semi-circle of infinite radius in the upper half-plane. However, as the integrand is odd in x (or, changing language to the conventions of complex analysis, in z), I think I can take the real part of an integral around a quarter-circle of infinite radius along the positive real axis and down the negative imaginary axis. This gives me a simple pole on the boundary at z=i\sqrt{t} and a simple pole at z=\sqrt{e^{i \pi/4}-t}. Plugging these in and taking the real part gives me
I(t)=Re \left[ \frac{\pi i}{4}e^{-i \pi /2}(e^{i \pi /4}-t)^{2} \right] =\frac{\pi}{4}(t^2-t\sqrt{2})
This is unfortunately wrong- mathematica gives me a decreasing function of t, but in a horribly complicated form, whilst I happen to know that a much simpler form exists. Any help would be greatly appreciated.
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