How to use residues with improper integrals?

  • Thread starter Thread starter laura_a
  • Start date Start date
  • Tags Tags
    Integrals
laura_a
Messages
64
Reaction score
0

Homework Statement



integral^infty_0 1/(x^2+1) (I know the answer (from text) is pi/2)


Homework Equations



Well that is what I need help with. I can see that there are 2 roots to the x^2+1 that are +/- i and I know from the text that I use the arc in the where x >= 0 so I use the z=i residue. I'm not sure what to do next in order to get the answer.

The Attempt at a Solution



I know that it is an even function so I can multiply the answer by 1/2 to get the value, and I also understand that the value of the integral is 2*pi*i *[Sum of residues]


Can anyone help me to understand. I have someone elses working out here on the same question and they are using limits and end up using ;im (z-i) = 1/2i
z->i (z-i)(z+i)

which I don't understand at all? How to get z-i and the other working out... it turns out that 1/2i is what I'm looking for because (1/2)2(pi)i(1/2i) is the answer... can anyone give me any extra info that might help me to understand, after this question I have to do the same thing but the question is dx/(x^2+1)^2 so I'm hoping that if I get help with the first one I'll understand the 2nd...

Thanks: :)
 
Physics news on Phys.org
Yes. You need to use the residue from the pole at z=i. Since 1/(z^2+1)=1/((z+i)*(z-i)), the residue is the value of this function after dropping the (z-i) factor evaluated at z=i. 1/(z+i) at z=i is 1/(2i). This is a good easy example to be clear about before you move on to harder ones.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top