So is there any way that I can reach you? I tried to use the methods in your video to evaluate the integral: ∫x^(1/3)/(x^4+1)dx from 0 to infinity. Of course I can deal with it and solve it by using other methods, but I want to see if I can solve it by the method you use in the video. However, when I try I cannot succes, so can you please help me? Best Anders
Use the semicircular contour in most cases. But if it is passing through the singularity, then use the Keyhole contour or the contour in Dirichlet Integral
it is really wonderfull video. it is not clear from your explanation why the integral around A and B converges to the required integral as r tends to 0 and R tends to 0. what is that you are using to justify this.i have solved the same problem but never thought why it converges.may be dominated convergence theorem may help us here.
Along the real axis, you want the integral to depart as little as possible from the original integral. If you use half a circle, your integral will traverse the real negative half-axis. and there´s no real square root there. Going (almost) full circle will give you the integral you want on the real positive half-axis.
@@gnydnnk8384 It´s a passage to the limit. The way you defined the branch cut, you never go across the positive real axis. Now, FIX a small r > 0, and a large R > 0. Because your arguments are in [0, 2pi), points z on A have (very small) positive arguments, theta(z), which are uniformly bounded above by r. For FIXED R, letting r go to zero makes angles vanish, and in that first limit you have the integral of your function between 0 and R. Now let R go to infinity, and the integral over A goes to your desired integral. Again, FIX a small r > 0, and a large R > 0. Because your arguments are in [0, 2pi), points z on B have (very large) positive arguments theta(z), very close to 2pi, but a little smaller than 2pi, which are uniformly bounded above by 2pi - r. For FIXED R, letting r go to zero makes angles go to 2pi, and in that first limit you have the integral of your function between 0 and R, multiplied by e^i(pi - 2pi) = e^(-i·pi) = -1. However, in doing so you are integrating ON THE FINITE SEGMENT [0, R] traversing it from left to right, whereas your motion along the closed circuit demands you traverse it from right to left, thus causing a second sign shift. All in all, for FIXED R you obtain, in the limit as r goes to zero, the integral of your original function from 0 to R. Now let R go to infinity, and the integral over B goes, again, to your desired integral. Therefore, IN THE LIMIT AS r goes to 0, and R goes to infty, the integrals over A (left to right) and B (right to left) tend to the same value. But you NEVER integrate over the positive half-axis, segments A are always ABOVE the (+) half-axis, segments B are always BELOW the (+) half-axis, the arguments of points of A are very small for very small r, and the arguments of points of B are very close to 2pi (but always smaller than 2pi) for very small r. Hope this helps.
