Finding Symmetric Poles for Complex Function Integrals

lonewolf5999
Messages
33
Reaction score
0
I'm looking for a function which has two simple poles, and whose integral along the positive real axis from 0 to infinity is equal to its integral along the positive imaginary axis.

I don't really know where to start. I'm looking at functions which have symmetry with respect to real/imaginary axes, i.e. if x is real, then f(x) = f(ix), which led me to consider things like f(z) = 1/(z^4) or f(z) = 1/(z^5), but those don't have simple poles and their integrals from 0 to infinity on the axes don't exist since they blow up at the origin. Any help is appreciated!
 
Physics news on Phys.org
Try locating your poles symmetrically with respect to the two axes. Like along the line x=y.
 
After trying your suggestion with z1 = 1 + i, z2 = -1 - i, and f(z) = 1/((z - z1)*(z-z2)), I ended up with the integral over the imaginary axis being the negative of that over the real axis, so I located my poles on the line y = -x instead, and that solved the problem.
Thanks for the help!
 
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...

Similar threads

Replies
17
Views
3K
Replies
5
Views
2K
Replies
27
Views
2K
Replies
3
Views
2K
Replies
9
Views
2K
Replies
2
Views
3K
Replies
14
Views
3K
Replies
3
Views
2K
Back
Top