I would like to evaluate some complicated integrals involving the hyperbolic secant, but the extension of the usual contour integration evaluation using the residue theorem isn't clear to me. I've been considering a simple example with a known solution $$\int_{-\infty}^{\infty} \text{Sech}\Big(\frac{\pi s}{2}\Big) \ d s \ = \ 2$$

This question is related as it discusses functions with an infinite number of poles, but there it is assumed that the sum of the residues converges. As is explained in that question, we are only ever actually considering bounded contours and then taking the limit that they grow arbitrarily large but I am not sure how that justifies my approach in this problem.

Suppose I am quite cavalier and I attempt to evaluate this integral by closing the contour along a semi-circle in the upper-half complex plane. Without really checking, I claim that the integral along the circular part of the contour vanishes and thus the integral over the real line is given by $$\sum_{n=0}^{\infty} 2 \pi i \ \text{Res}\Big( \text{Sech}\Big(i\pi\big (n+\frac{1}{2}\big)\Big) = 4-4+4-4 \ . . .$$ which clearly does not converge. However, if we consider the average of the first $m$ partial sums $$\frac{1}{m}(4+0+4+0+ \ . . . ) = \left \{ \begin{array}{lr} 2 & \ m \in \text{evens}\\ 2 + \frac{4}{m}& m \in \text{odds} \end{array} \right.$$ which goes to $2$ as $m \to \infty$. This procedure seems totally ad hoc, even magical, but it predicts the correct answer. The first question is, why does the limit of the average of the partial sums give the correct answer?

Another way of regularizing this problem is to instead compute $$\lim_{\eta \to 0}\int_{-\infty}^{\infty} \text{Sech}\Big(\frac{\pi s}{2}\Big) e^{i \eta \ s} \ d s $$ and then using the same procedure $$\lim_{\eta \to 0}\sum_{n=0}^{\infty} 2 \pi i \ \text{Res}\Big( \text{Sech}\Big(i\pi\big (n+\frac{1}{2}\big) e^{-\eta (2n-1)}\Big) =\lim_{\eta \to 0} \frac{4e^{\eta}}{1+e^{2\eta}} = 2$$ also gives the correct answer. So, there seems to be something to this method.

If I am more careful about the circular piece of the contour, I can show that its contribution is exponentially small almost everywhere. $\text{Sech}(x + iy)$ clearly decreases exponentially when $x$ is large, but what about when $y$ is large and $x$ is small? By rewriting $x + iy = re^{i \theta}$ and taking the absolute value, I can show that when $\theta = \frac{\pi}{2} +\epsilon, \ \left|\epsilon\right|<1$ (ie we are close to the imaginary axis), then $$\left| \text{Sech}(re^{i\theta})\right|\sim e^{-\frac{1}{2}\pi r \left|\epsilon\right| } : \ \ r\to \infty$$ My understanding of the Riemann integral suggests that the single point where $\epsilon = 0$ does not change the value of the integral along the contour.

So, it seems like I do have good reason to argue that the contribution from the circular contour vanishes. Is there a way, then, to justify my regularized sum of the residues as the correct answer in general or is it just an accident for this special case?

  • $\begingroup$ Pick any small positive number $\delta > 0$. Based on your estimate, the range of $\theta$ such that $\left|\frac{1}{\cosh(re^{i\theta})}\right| > \delta$ can be as big as $O(\frac{4|\log\delta|}{\pi r})$. Since $|dz| = |r d\theta|$, the contribution from the piece you excluded can be as big as $O(\frac{4\delta|\log\delta|}{\pi})$. Since this is independent of $r$, you cannot conclude the contribution from the circular contour vanishes. $\endgroup$ – achille hui Sep 25 '13 at 5:02
  • $\begingroup$ @achillehui I think I follow your logic. Can you explain then why I seem to get the correct answer? An accident? $\endgroup$ – Kevin Driscoll Sep 25 '13 at 5:29
  • $\begingroup$ Ultimately, this comes down to the periodic property of cosh: $\cosh(x+i k\pi) = (-1)^k\cosh(x)$. This is also the key factor behind Ron's answer. $\endgroup$ – achille hui Sep 25 '13 at 5:38
  • $\begingroup$ @achillehui Hmmmmmm, I am not sure how that applies to what I have done, but I will think about it some more $\endgroup$ – Kevin Driscoll Sep 25 '13 at 5:41

Of course we can avoid contours altogether, $$ \begin{align} \int_{-\infty}^\infty\mathrm{sech}(ax)\,\mathrm{d}x &=4\int_0^\infty\frac{e^{-ax}\,\mathrm{d}x}{1+e^{-2ax}}\\ &=4\int_0^\infty\left(e^{-ax}-e^{-3ax}+e^{-5ax}-\dots\right)\,\mathrm{d}x\\ &=\frac4a\left(1-\frac13+\frac15-\dots\right)\\ &=\frac4a\frac\pi4\\ &=\frac\pi{a}\tag{1} \end{align} $$ However, the integral along the contours with large imaginary part cannot be discounted since $$ \cosh(x+iy)=\cosh(x)\cos(y)+i\sinh(x)\sin(y)\tag{2} $$ That is, for any $k\in\mathbb{Z}$, $$ \int_{-\infty}^\infty\mathrm{sech}(ax+ik\pi)\,\mathrm{d}x=(-1)^k\frac\pi{a}\tag{3} $$ The method outlined by Ron Gordon, where we stick close to the real axis, is a good way to approach integrals of this kind.

Why does the average give the correct answer?

Consider the contour $$ \gamma_{k,R}=[-R,R]\color{#A0A0A0}{\cup[R,R+ik\pi]}\cup[R+ik\pi,-R+ik\pi]\color{#A0A0A0}{\cup[-R+ik\pi,-R]}\tag{4} $$ The integrals along the greyed out pieces vanish as $R\to\infty$. Thus, using $(2)$, $$ \begin{align} \lim_{R\to\infty}\int_{\gamma_{k,R}}\mathrm{sech}(x)\,\mathrm{d}x &=\int_{-\infty}^\infty\mathrm{sech}(x)\,\mathrm{d}x -\int_{-\infty+ik\pi}^{\infty+ik\pi}\mathrm{sech}(x)\,\mathrm{d}x\\ &=\int_{-\infty}^\infty\mathrm{sech}(x)\,\mathrm{d}x -(-1)^k\int_{-\infty}^\infty\mathrm{sech}(x)\,\mathrm{d}x\\ &=\left(1-(-1)^k\right)\int_{-\infty}^\infty\mathrm{sech}(x)\,\mathrm{d}x\tag{5} \end{align} $$ Thus, the fact that the average converges to the proper thing is nothing more than the fact that the average of $$ \left(1-(-1)^k\right)\tag{6} $$ tends to $1$.

What I have said above could be made into a valid argument that the average gives the proper value. However, you could just stop at $k=1$ and say that that gives you twice the actual value and divide by $2$. This is just what Ron Gordon does.

Concerning $e^{\large-\frac12\pi r|\epsilon|}$

As noted in $(2)$, $$ \begin{align} |\mathrm{sech}(x+iy)| &=\left|\,\frac1{\cosh(x)\cos(y)+i\sinh(x)\sin(y)}\,\right|\\ &\sim2e^{-|x|}\\ &=2e^{-r\,|\sin(\theta)|}\\ &\sim2e^{-r\,|\theta|}\tag{7} \end{align} $$ where $\theta$ is the angle from the imaginary axis. This estimate is deceptive. Although it is true, inside the strip where $2e^{-r\,|\theta|}\ge1$, the integral along the arc is approximately $$ \int_{-\log(2)}^{\log(2)}\mathrm{sech}(x+iy)\,\mathrm{d}x\tag{8} $$ which can blow up when $y\equiv\frac\pi2\pmod{\pi}$, but whose absolute value is always at least $\frac85\log(2)$ (e.g. when $y\equiv0\pmod{\pi}$).

  • $\begingroup$ Your decomposition seems like a reasonable argument that the circular contour cannot be neglected. But I thought I showed that it can. The piece I seem to be missing is how your objection and @achillehui objection connect and show that my argument was incorrect. That is, you both seem to be saying that I accidentally got the right answer but I don't see why. $\endgroup$ – Kevin Driscoll Sep 25 '13 at 7:09

There is a much easier way to use contour integration to evaluate the integral: use a different contour. I recommend a rectangular contour $C$, with vertices: $\{-R,R,R+2 i,-R+2 i\}$. That way, we have only one pole $z=i$ inside $C$. Thus:

$$\oint_C \frac{dz}{\cosh{\frac{\pi}{2} z}} = \int_{-R}^R \frac{dx}{\cosh{\frac{\pi}{2} x}} + i \int_0^2 \frac{dy}{\cosh{\frac{\pi}{2} (R+i y)}} \\+\int_{R}^{-R} \frac{dx}{\cosh{\frac{\pi}{2} (x+2 i)}} + i \int_2^0 \frac{dy}{\cosh{\frac{\pi}{2} (-R+i y)}} $$

The second and fourth integrals vanish as $R \to \infty$. The first and third integrals are equal, as $\cosh{\frac{\pi}{2} (x+2 i)} = -\cosh{\frac{\pi}{2} x}$, and we reverse the direction of the third integral.

On the other hand, the contour integral is $i 2 \pi$ times the residue at the pole $z=i$, which is $-i 2/\pi$. (I'll leave that to the reader.) Then we have

$$2 \int_{-\infty}^{\infty} \frac{dx}{\cosh{\frac{\pi}{2} x}} = i 2 \pi \left ( -i \frac{2}{\pi}\right ) = 4$$

The sought-after result follows.

  • $\begingroup$ Thanks Ron for your answer. I am not sure that the more complicated integrals I need to do can make use of the clever periodicity that you used, but perhaps it is possible. $\endgroup$ – Kevin Driscoll Sep 25 '13 at 5:26
  • $\begingroup$ @KevinDriscoll: the integrals of many oscillatory functions may be attacked with variants of rectangular contours. The main point is to find an integration path from which you may recover the original integral while having integral vanish along other paths. There is a little bit of an art to it, but it is a well-trodden path. $\endgroup$ – Ron Gordon Sep 25 '13 at 5:28
  • $\begingroup$ It's a good strategy that I hadn't thought of. I think it will work for one integral that contains only sinh's and sech's, but another has a factor of $\frac{s \sinh{\frac{\pi}{2}s}}{s \sinh{\frac{\pi}{2}s}-1}$ where it isn't obvious that there's a periodic path. I'll investigate some more and perhaps post the whole integral if I can't see anything. $\endgroup$ – Kevin Driscoll Sep 25 '13 at 6:03
  • $\begingroup$ @KevinDriscoll: contour methods - or any other methods - cannot get around nasty integrands, such as those having a combination of monomials and hyperbolic/exponential functions. I doubt the shape of your contour can change that. $\endgroup$ – Ron Gordon Sep 25 '13 at 12:16

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