Ron Gordon
Reputation
381/400 score
 2h comment Trying to evaluate an infinite sum using method of residues math.stackexchange.com/questions/359667/… 16h revised Integrate $\int \frac{\arctan\sqrt{\frac{x}{2}}dx}{\sqrt{x+2}}$ added 7 characters in body 17h answered Integrate $\int \frac{\arctan\sqrt{\frac{x}{2}}dx}{\sqrt{x+2}}$ 17h revised Recursive integral definition for the dynamics of a DC motor deleted 1 character in body 20h comment Recursive integral definition for the dynamics of a DC motor As I said, there are many ways to solve such first-order equations. For one, you can consider $$\frac{d}{dt}(r e^{-D t})$$. 21h answered Recursive integral definition for the dynamics of a DC motor 1d revised Calculating convolution integral analytically added 1 character in body 2d comment Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$ Jack, with you I figured it was something as trivial as that. 2d comment Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$ Jack, no offense, but your previous answer was wrong and your correction was merely a change to the correct answer provided by others. I think you really should show correction to a step as well. For example, are you sure your partial fractions decomposition is correct? Hint: @Leg got it right. 2d revised Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$ edited tags 2d answered Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$ 2d comment What happens to poles lying on branch cuts in contour integration? @Taozi; yes, that's right. The fact that the sqrt is in the denominator makes those integrals cancel. The contribution of the arc about the origin will vanish so long as the power of $z$ in the denominator is less than $1$. Nov 25 comment What happens to poles lying on branch cuts in contour integration? @Taozi: thanks for the kind words. Regarding the combination of the 2nd and the 4th integrals, that is what becomes the Cauchy principal value of the integral over combined region. This is specifically designed to deal with the pole at $x=1$. Note that $$PV \int_0^{\infty} dx \frac{e^{t x}}{\sqrt{x} (1-x)} = \lim_{\epsilon \to 0} \left [\int_0^{1-\epsilon} dx \frac{e^{t x}}{\sqrt{x} (1-x)} + \int_{1+\epsilon}^{\infty} dx \frac{e^{t x}}{\sqrt{x} (1-x)} \right ]$$ I hope this helps. Nov 25 revised What happens to poles lying on branch cuts in contour integration? added 34 characters in body Nov 25 revised What happens to poles lying on branch cuts in contour integration? edited tags Nov 25 revised What happens to poles lying on branch cuts in contour integration? added 22 characters in body Nov 25 answered What happens to poles lying on branch cuts in contour integration? Nov 24 comment Fourier transform of a test function Maybe you might want to let one of us know what exactly is a test function. Nov 24 comment Determine residues of $\frac{e^{-\sqrt{z(z+r))}}}{1+\alpha\sqrt{z(z+r)} + (1-\alpha \sqrt{z(z+r)})e^{(-\sqrt{z(z+r)})}}$ How is this limit even related to a pole in the given function? Plugging in $\alpha=0$ does not produce a zero in the denominator. Nov 24 comment Trying to evaluate an improper integral using the methods of complex analysis $z = R e^{i \theta} \implies dz = i R e^{i \theta} d\theta$.