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Suppose the function $f(z)$ has singularities and I want to calculate the integral $$\int_{-\infty}^{\infty}f(z)dz=?$$

I use Residue theorem

$$\oint_cf(z)dz=2\pi i\sum\text{Res}(f,z_0)$$

And

$$\oint_cf(z)dz=\int_{-\infty}^{\infty}f(z)dz+\int_{arc}f(z)dz$$

Quenstion: In general, is the arc part always zero? How to calculate the arc part?

$$\int_{arc}f(z)dz=?$$

If necessary, pls show a specific example.

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  • $\begingroup$ The arc integral need not equate to zero. With the right contour however, the arc integrals usually turn into something much more easily managed. $\endgroup$ – Frank W. Jan 26 at 23:03
  • $\begingroup$ Can you elaborate? How exactly to calculate the arc part? $\endgroup$ – kinder chen Jan 27 at 0:39
  • $\begingroup$ @Kinder: your question is ill-posed, and your last question is more or less like how to calculate a generic integral? Hard to say, if we do not restrict our attention to a class of somehow-well-behaving functions. For instance, the integral of $\frac{\sin z}{z}$ over $\{\text{Im}(z)>0,|z|=R\}$ does not converge to zero as $R\to +\infty$, but the integral of $\frac{e^{iz}}{z}$ over the same region does. $\endgroup$ – Jack D'Aurizio Jan 27 at 5:30
  • $\begingroup$ @JackD'Aurizio, sorry for the bothering. I'm a beginner, I don't know what kind of function I should show. I have no clue how to calculate the arc part. I thought it would converge to 0. $\endgroup$ – kinder chen Jan 27 at 6:23

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