The following is an integral I am trying to evaluate

$$I= \int_{-\infty}^\infty f(s) \, ds = \int_{-\infty}^\infty \frac{\frac{1}{(1- \ \ 2 \pi j s )^{m}}-1}{2\pi j s }\ e^{-2\pi j s \ \theta}\ ds $$

where $\theta$ is non-negative constant and $m$ is an positive integer.

Someone helped me by saying that, I can solve it exactly by closing the integration contour in the lower half complex plane, i.e using the Residue formula. Therefore I tried that below

What I tried

Using the Residue Theorem "Residue Theorem" and knowing that the pole is at $z^*=\frac{1}{2\pi j}$ then

$$I = -2 \pi j \ \ \text{Res}_{z^*= \frac{1}{2\pi j}}[f(z)]$$

Next I evaluate the residue "residue of function", then

\begin{align}\text{Res}_{z^*= \frac{1}{2\pi j}}\left[f(z)\right]=&\lim_{z\rightarrow z^*} (z-z^*) \frac{\frac{1}{(1- \ \ 2 \pi j z )^{m}}-1}{2\pi j z }\ e^{-2\pi j z\theta}\\ \\ &=\lim_{z\rightarrow z^*}\ (z-z^*) \frac{\frac{1}{(1- \frac {z}{z*} )^{m}}-1}{2\pi j z }\ e^{-2\pi j z \theta}\\ &=\lim_{z\rightarrow z^*} (z-z^*) \frac{\frac{(z^*)^m}{(z^*-{z} )^{m}}-1}{2\pi j z }\ e^{-2\pi j z\theta}\\ &=\lim_{z\rightarrow z^*}\frac{\frac{- (z^*)^m}{(z^*-{z} )^{m-1}}-1}{2\pi j z }\ e^{-2\pi j z\theta}\\ \\ &= ???? \end{align}

I don't know if I am doing the right thing, do you think my derivation is correct? Am I at least on the right track?


  • 2
    $\begingroup$ You're using the formula for the residue at a simple pole, but you have a pole of order $m$. Write $f(z) = \frac{g(z)}{(z - z^\ast)^m}$, where $$g(z) = (-1)^m\frac{1-(1-2\pi jz)^m}{(2\pi j)^m(2\pi j z)}e^{-2\pi j z\theta},$$ then the residue is $$\frac{1}{(m-1)!}\left(\frac{d}{dz}\right)^{m-1}\biggl\lvert_{z=z^\ast} g(z).$$ $\endgroup$ Dec 17, 2014 at 20:56
  • $\begingroup$ Thanks Daniel, I didn't understand your final formula do you mean this instead, $\frac{1}{(m-1)!}(\frac{d}{dz})^{m-1} g(z) \big|_{z=z^*}$? $\endgroup$
    – Henry
    Dec 17, 2014 at 21:00
  • $\begingroup$ @Henry You have to derive $m-1$ times. $\endgroup$
    – Ant
    Dec 17, 2014 at 21:03
  • $\begingroup$ No, you need to differentiate $m-1$ times, not just once (and if $m = 1$, you don't differentiate at all). Expand $g$ into its Taylor series about $z^\ast$, then you'll see that you need the coefficient of $(z-z^\ast)^{m-1}$. $\endgroup$ Dec 17, 2014 at 21:03
  • $\begingroup$ @Daniel Fisher, thanks again, so if $m=2$, then i take the derivative of $g(z)$ twice let us call it $g{''}(z)$, then evaluate at $z=z^*$. $\endgroup$
    – Henry
    Dec 17, 2014 at 21:06

1 Answer 1


You are using the formula for the residue in the case of a simple pole, but here we have a pole of order $m$ (which may be $1$, but generally isn't), so the formula to use is

$$\operatorname{Res}_{z^\ast = \frac{1}{2\pi j}} [f(z)] = \lim_{z\to z^\ast} \frac{1}{(m-1)!}\biggl(\frac{d}{dz}\biggr)^{m-1} (z-z^\ast)^mf(z).\tag{1}$$

When we write $f$ in the form

$$f(z) = \frac{g(z)}{(z-z^\ast)^m},$$

a Taylor expansion of $g$ shows that

$$f(z) = \sum_{n=0}^\infty \frac{g^{(n)}(z^\ast)}{n!}\cdot (z-z^\ast)^{n-m},$$

and the residue is the coefficient of $(z-z^\ast)^{-1}$, so the term for $n = m-1$, which is


Here, we have

$$f(z) = \frac{\frac{1}{(1-2\pi jz)^m}-1}{2\pi j z}e^{-2\pi jz\theta} = \frac{\frac{1-(1-2\pi jz)^m}{2\pi jz}}{(1-2\pi jz)^m}e^{-2\pi j z\theta} = \frac{(-1)^m\frac{1-(1-2\pi jz)^m}{(2\pi j)^m(2\pi jz)}e^{-2\pi jz\theta}}{\bigl(z-\frac{1}{2\pi j}\bigr)^m},$$


$$g(z) = (-1)^m\frac{1-(1-2\pi jz)^m}{(2\pi j)^m(2\pi jz)}e^{-2\pi jz\theta}.$$

Since $(1-2\pi jz)^m$ has a zero of order $m$ at $z^\ast = \frac{1}{2\pi j}$, that part of $g$ doesn't contribute to the residue, and we only need to compute the derivatives of

$$\frac{1}{2\pi jz} e^{-2\pi j z\theta}$$

and multiply with the constant factors, which makes the computation a bit simpler.

  • $\begingroup$ Does this method work if $m$ is not an integer? $\endgroup$
    – Henry
    Dec 17, 2014 at 22:10
  • $\begingroup$ No, if $m$ is not an integer, then your integrand doesn't have a pole at $z^\ast$ but a branch-point, and you must have a branch-cut from $z^\ast$ to $\infty$. Sometimes, one can use branch-cuts and the residue theorem to compute integrals (things like $\int_0^\infty \frac{\log x}{1+x^2}\,dx$), but I don't see a way to do that here for non-integer $m$. $\endgroup$ Dec 17, 2014 at 22:35
  • $\begingroup$ if you don't mind me commenting again on this, i was wondering why isnt the pole at zero considered in the analysis? thanks $\endgroup$
    – Henry
    Feb 11, 2015 at 12:23
  • $\begingroup$ Here we don't have a pole at $0$, the zero of the denominator $2\pi js$ is cancelled by the zero of the numerator $\frac{1}{(1-2\pi js)^m}-1$ at $0$, so the integrand is analytic in a neighbourhood of $0$ [after removing the removable singularity at $0$]. $\endgroup$ Feb 11, 2015 at 12:31
  • $\begingroup$ is this the same case as the integrand i mentioned in this question I asked here math.stackexchange.com/questions/1142633/… Reason I am asking is because I get comments that zero is also a pole...@Daniel Fischer $\endgroup$
    – Henry
    Feb 11, 2015 at 14:45

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