How to Solve this Tough Integration Question with LaTex Commands?

  • Thread starter Thread starter trap
  • Start date Start date
  • Tags Tags
    Integration
trap
Messages
51
Reaction score
0
Anyone know how to solve this?
\text 8\pi\int_{0}^{\infty}\frac{t^3}{(4+t^2)^\frac{5}{2}} dt
 
Physics news on Phys.org
Sure, make the following substitution:

u = 4 + t^2[/itex]
 
yeah, i tried that, but i can't go further coz i can't get rid of the t^3
 
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:
can you help me with one more step? i really can't make two integrals with power of u..
 
ok, i got it now, thanks for the help
 
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.
 
can't it be done by contour integration?
 
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
Nice. Here we go: (just do a click on the equation and a pop-up window will display the LaTex commands)

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