# Tough Integration question

1. Oct 3, 2005

### trap

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

2. Oct 3, 2005

### hotvette

Sure, make the following substitution:

$$u = 4 + t^2[/itex] 3. Oct 3, 2005 ### trap yeah, i tried that, but i can't go further coz i can't get rid of the t^3 4. Oct 3, 2005 ### hotvette No need to get rid of it. Take it further. I believe you'll end up with 2 integrals in fractional powers of u. Last edited: Oct 4, 2005 5. Oct 3, 2005 ### trap can you help me with one more step? i really can't make two integrals with power of u.. 6. Oct 3, 2005 ### trap ok, i got it now, thanks for the help 7. Oct 4, 2005 ### dextercioby HINT: [tex] \sqrt{4+t^{2}}=u$$

$$t \ dt= u \ du$$

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

Daniel.

8. Oct 4, 2005

### murshid_islam

can't it be done by contour integration?

9. Oct 5, 2005

### 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.

10. Oct 5, 2005

### 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}