Tough Integration question

[trap]

Anyone know how to solve this?
\text 8\pi\int_{0}^{\infty}\frac{t^3}{(4+t^2)^\frac{5}{2 }} dt

[hotvette]

Sure, make the following substitution:

[tex]u = 4 + t^2[/itex]

[trap]

yeah, i tried that, but i can't go further coz i can't get rid of the t^3

[hotvette]

No need to get rid of it. Take it further. I believe you'll end up with 2 integrals in fractional powers of u.

[trap]

can you help me with one more step? i really can't make two integrals with power of u..

[trap]

ok, i got it now, thanks for the help

[dextercioby]

HINT:

\sqrt{4+t^{2}}=u

t \ dt= u \ du

\int_{0}^{\infty} \frac{t^{3}+4t-4t}{\left(\sqrt{t^{2}+4}\right)^{5}} \ dt =\int_{0}^{\infty} \frac{t^{2}+4}{\left(\sqrt{t^{2}+4}\right)^{5}}t \ dt - 4\int_{0}^{\infty}\frac{t \ dt}{\left(\sqrt{t^{2}+4}\right)^{5}}

Daniel.

[murshid_islam]

can't it be done by contour integration?

[benorin]

Let t^2=x-4 -> 2tdt=dx and 0<=t<=infinity -> 4<=x<=infinity

so the integral becomes 4*Pi*Int((x-4)*x^(-5/2),x=4..infinity) = 8*Pi/3

excuse my maple notation, I have yet to learn LaTeX.

[saltydog]

Nice. Here we go: (just do a click on the equation and a pop-up window will display the LaTex commands)


\begin{align*}
8\pi\int_0^{\infty}\frac{t^3}{(4+t^2)^{5/2}}dt &=
\frac{8\pi}{2}\int_4^{\infty}\frac{x-4}{x^{5/2}}dx \\ &=
4\pi\int_4^{\infty}x^{-5/2}(x-4)dx \\ &=
4\pi\int_4^{\infty}(x^{-3/2}-4x^{-5/2})dx \\ &=
4\pi\left(8/3x^{-5/2}-2x^{-1/2}\right)_4^{\infty} \\ &=
4\pi\left(-(1/3-1)\right) \\ &=
\frac{8\pi}{3}
\end{align}