Contour integral Definition and 116 Threads

so suppose i wanted to calculate the antiderivatives of e^x\sin{x} and just for the hell of it also e^x\cos{x}. well i could perform integration by parts twice recognize that the original integral when it reappears, subtract from one side to the other blah blah blah. or i could pervert a...

We have an integral over q from -\infty to +\infty as a contour integral in the complex q plane. If the integrand vanishes fast enough as the absolute value of q goes to infinity, we can rotate this contour counterclockwise by 90 degrees, so that it runs from -i\infty to +i\infty. In making...

Can anyone recommend a good introduction to contour integrals for someone not taking complex analysis? We are doing these integrals in a physics class and I'm terribly confused. I know that I have to choose contours that "go around" my poles, but I don't understand how to do this (I can't seem...

Using Cauchy's integral theorem how could we compute \oint _{C}dz D^{r} \delta (z) z^{-m} since delta (z) is not strictly an analytic function and we have a pole of order 'm' here C is a closed contour in complex plane

I have two related questions. First of all, we have the identity: \int_{-\infty}^{\infty} e^{ikx} dk = 2 \pi \delta(x) I'm wondering if it's possible to get this by contour integration. It's not hard to show that the function is zero for x non-zero, but the behavior at x=0 is bugging...

Homework Statement Find the contour integral of e(iqz)/z^4 The Attempt at a Solution Is it infinite as the pole at z=0 is too high?

I'm trying to find \int_{-\infty}^{\infty} \frac{exp(ax)}{cosh(x)} dx where 0<a<1 and x is taken to be real. I'm doing this by contour integration using a contour with corners +- R, +- R + i(pi), and I'm getting an imaginary answer which is \frac{2i\pi}{sin (a \pi)}. I'm thinking this is...

Contour integral How would you deal with this? \int \frac{\rho \sin{\theta} d \rho d \theta}{\cos{\theta}} \frac{K^2}{K^2 + \rho^2} e^{i \rho \cos{\theta} f(\mathbf{x})} if the cos(theta) were'nt on the bottom I'd have no problem; I'd simply substitute for cos(theta) and the sin(theta)...

\int \frac{\rho^4 \sin^3{\theta} d \rho d \theta e^{i \rho r \cos{\theta}}}{(2 \pi)^2 [K^2 + \rho^2]} I am confused about where the singularities are in this function. Will they simply be at \rho = iK and -ik or does the \rho^4 factor make a difference? Also the sin^3(\theta) e^(i \rho cos...

I was w=kind of confused as of how to go about solving this integeal using complex methods. it is the Integral from 0 to infinity of{dx((x^2)(Sin[xr])}/[((x^2)+(m^2))x*r] where m and r are real variables. I tried to choose a half "donut" in the upper part of the plane with radii or p and R...

For my homework I am told: "Evaluate $z^(1/2)dz around the indicated not necessarily circular closed contour C = C1+C2. (C1 is above the x axis, C2 below, both passing counter-clockwise and through the points (3,0) and (-3,0)). Use the branch r>0, -pi/2 < theta < 3*pi/2 for C1, and the branch...

How might one evaluate an integral equation like the following: I = lim k-> 0+ {ClosedContourIntegral around y [1/(z^2 + k^2)]}, where the contour y is a simple, closed, and positively oriented curve that encloses the simple pole at z = i*K? Is it possible to evaluate integrals of this...

is this the right section to post contour integral questions?

Hi, I've typed up my work. Please see the attached pdf. Basically, I am trying to sovle this problem. \int_0^\infty \frac{x^\alpha}{x^2+b^2} \mathrm{d}x for 0 <\alpha < 1. I follow the procedure given in Boas pg 608 (2nd edition)...and everything seems to work. However, when I...

Hi, I'm having a bit of trouble with this question. Use the property |integral over c of f(z)dz|<=ML to show |integral over c of 1/(z^2-i) dz|<=3pi/4 where c is the circle |z|=3 traversed once counterclockwise thanks in advance for any tips.

I'm trying to solve this contour integral shown on the attached file, I know usually that they involve curved lines. I know that this is trivial but I need some help with the problem. Please take a look.