Multi-keyhole contour integral with branch cut

Question

How to construct a contour to calculate complex line integral

$$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\arctan\left(\frac{1}{z}\right)\arctan\left(\frac{1}{s-z}\right)\,\mathrm{d}z$$

This integral is derived from the Laplace transform of the square of the Sinc function.

$$\mathscr{L}[f^2(t);s]=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}F\left(z\right)F\left(s-z\right)\,\mathrm{d}z$$

As shown in the expression above, Wikipedia gives the Laplace transform formula of function product as. I want to verify this formula by taking function as the Sinc function. I've found six singularities $z=\pm\,\!i,0,s,s\pm\,\!i$ so far, but I don't know how to construct the contour.

I already know how to calculate $\displaystyle\int_{0}^{+\infty}\!\left(\frac{\sin(x)}{x}\right)^{\!\!2}\,e^{-sx}\,dx$ through the parametric integral, but now I don't know how to construct the contour and use the residue theorem to calculate the arctangent complex integral.

Answer 1

The key to the business is that the line $c-i\infty$ to $c+i\infty$ (or in the limit if you like) must be completely within the region of convergence of $F(z)=\arctan(1/z)=\operatorname{arccot}(z).$ We also certainly don't want to run afoul of the singularities of $\arctan(1/(s-z))=\operatorname{arccot}(s-z).$ The singularities you have mentioned are the only ones we need to concern ourselves with. Perhaps the easiest way to choose $c$ is to make $c>\max(0,\operatorname{Re}(s)),$ but the problem with that is that if we chose a D-shaped contour with the half-circle off to the right, the Residue Theorem would give us zero for the integral as well as for the portion on the half-circle. We have to include at least one singularity inside the contour in order to make everything work for us. So let us assume that $\operatorname{Re}(s)>0,$ and then we need to choose $c: 0<c<\operatorname{Re}(s).$ Thus, if we choose a D-shaped contour now, with the half-circle off to the right, we will pick up the three singularities at $s, s\pm i.$ Just to be precise, let us choose $\Gamma=\gamma_1+\gamma_2$ as the full contour, where: \begin{align*} \gamma_1&:c+it,\;t\in[-R,R]\\ \gamma_2&:c+Re^{-i\theta},\;\theta\in[-\pi/2,\pi/2] \end{align*} Then we would have the integral over $\gamma_1$ as being the integral of interest, the integral over $\gamma_2$ we hope to be zero (maybe by the $ML$ lemma), so that the Residue Theorem can give us the final result. Let $$f(z)=\arctan\!\left(\frac1z\right)\arctan\!\left(\frac{1}{s-z}\right).$$ The Residue Theorem gives us that $$\frac{1}{2\pi i}\int_\Gamma f(z)\,dz=\sum_{k=1}^3\operatorname{Res}(f,a_k),$$ where \begin{align*} a_1&=s-i\\ a_2&=s\\ a_3&=s+i \end{align*} are our three singularities. Now the $ML$ lemma will indeed show us that $$\int_{\gamma_2}f(z)\,dz=0$$ in the limit, since the arguments of both $\arctan$ functions will go to zero; these functions go to zero faster than $R$ gets big. So we can now write $$\frac{1}{2\pi i}\int_{\gamma_1}f(z)\,dz=\sum_{k=1}^3\operatorname{Res}(f,a_k).$$ So now you need to evaluate the residues.

One possible "gotcha": the kind of singularity of $f(z)$ at $s.$ If that singularity gives you problems, then you can change the contour to avoid that singularity by skirting it like this. First of all, you would set $c=\operatorname{Re}(s).$ Then you do: \begin{align*} \gamma_1&:c+it,\;t\in[-R,\operatorname{Im}(s)-1-\varepsilon]\\ \gamma_2&:s-i+\varepsilon e^{-i\theta},\;\theta\in[-3\pi/2,-\pi/2]\\ \gamma_3&:c+it,\;t\in[\operatorname{Im}(s)-1+\varepsilon,\operatorname{Im}(s)-\varepsilon]\\ \gamma_4&:s+\varepsilon e^{i\theta},\;\theta\in[-\pi/2,\pi/2]\\ \gamma_5&:c+it,\;t\in[\operatorname{Im}(s)+\varepsilon,\operatorname{Im}(s)+1-\varepsilon]\\ \gamma_6&:s+i+\varepsilon e^{-i\theta},\;\theta\in[-3\pi/2,-\pi/2]\\ \gamma_7&:c+it,\;t\in[\operatorname{Im}(s)+1+\varepsilon,R]\\ \gamma_8&:c+Re^{-i\theta},\;\theta\in[-\pi/2,\pi/2]. \end{align*} There's a lot here, but it comes down to skirting $s\pm i$ to the left so as to include those singularities, but to exclude $s$ itself by skirting to the right. The hope is that $\displaystyle\int_{\gamma_2}\to 0$ as $\varepsilon\to 0,$ as well as $\displaystyle\int_{\gamma_4}$ and $\displaystyle\int_{\gamma_6}.$ In this case, you would have $$\int_{\gamma_1}+\int_{\gamma_3}+\int_{\gamma_5}+\int_{\gamma_7}=2\pi i(\operatorname{Res}(f,a_1)+\operatorname{Res}(f,a_3)).$$

Answer 2

We want to evaluate $\displaystyle I(S)=\int_{0}^{+\infty}\!\left(\frac{\sin(x)}{x}\right)^{\!\!2}\,e^{-sx}\,dx$ via contour integration. In order to ensure that our result is correct, let's first evaluate the integral using an easier method. After that we will accomplish a complex integration program.

$$\frac{\partial^2}{\partial s^2}I(s)=\int_0^\infty\sin^2xe^{-sx}dx=\frac{1}{2s}-\frac{1}{2}\Re \int_0^\infty e^{2ix-sx}dx=\frac{1}{2s}-\frac{1}{2}\frac{s}{s^2+4}$$ Integrating two times $$I(s)=\frac{s}{2}\,\log s-\frac{s}{4}\,\log(s^2+4)-\frac{i}{2}\log\frac{s+2i}{s-2i}+As+B$$ where $A$ and $B$ are some constants to be defined.

We will also use $\tan^{-1}(z)\,$ representation: $$\tan^{-1}(z)=-\frac{i}{2}\log\frac{i+z}{i-z}=-\frac{i}{2}\log\frac{1+iz}{1-iz}$$

$\text{At}\,s\to\infty\,\,\, I(s)\to 0 \,\,\Rightarrow\,\,\, A=B=0$

$\text{At}\,s\to0\,\,\, I(s)\to \int_0^{\infty}\frac{\sin^2x}{x^2}dx=\frac{\pi}{2}\,\,\Rightarrow\,\, -\frac{i}{2}\log\frac{s+2i}{s-2i}\to \frac{\pi}{2}$, and we should interpret

$$-\frac{i}{2}\log\frac{s+2i}{s-2i}=-\frac{i}{2}\log\big(-\frac{i+s/2}{i-s/2} \big)=\frac{\pi}{2}-\tan^{-1}\Big(\frac{s}{2}\Big)\to 0 \,\text{at} \,\,s\to 0\,$$

Finally, $$I(s)=\frac{s}{2}\,\log s-\frac{s}{4}\,\log(s^2+4)+\frac{\pi}{2}-\tan^{-1}\Big(\frac{s}{2}\Big)$$

---

Now, let's get this result via integration in the complex plane. $$I(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}F\left(z\right)F\left(s-z\right)\,\mathrm{d}z,\,\,\text{where}\,\, F(z)=\int_0^\infty\frac{\sin x}{x}e^{-xz}dx$$ It is not difficult to find $\,F(z)=\tan^{-1}(\frac{1}{z})=\frac{i}{2}\log\frac{z-i}{z+i}\,$, and $\,\,F(s-z)=\frac{i}{2}\log\frac{z-s+i}{z-s-i}$

At $|z|\to\infty \,\,\,F(z)F(s-z)\,\sim\frac{1}{4}\log\frac{1+i/z}{1-i/z}\log\frac{1+(s-i)/z}{1+(i-s)/z}\sim\frac{i(s-i)}{z^2}$

The integrand declines rapidly enough, and we can close the contour by a segment of a big circle with radius $R$; integral along this circle does not contribute at $R\to\infty$. $$I(S)=-\frac{1}{8\pi i}\int_{c-i\infty}^{c+i\infty}\log\frac{z-i}{z+i}\log\frac{z-s+i}{z-s-i}dz=-\frac{1}{8\pi i}\oint\log\frac{z-i}{z+i}\log\frac{z-s+i}{z-s-i}dz$$ We should note that this formula is applicable for $c>p$ and $s>p+c$, where $p$ is the growth power for $f(z)$. For $\,\frac{\sin x}{x} \,\,\,p=0$, so the line of integration in the complex plane goes between the points $z=0$ and $z=s$. This line position can be also defined by the requirement for the integral convergence for any $s>0$.

We can close the contour by the circle segment either in the right or left side of the complex plane. Let's choose the left side (you can check that closing the contour in the right side gives the same result). To make the integrand single-valued we have to add cuts connecting branch points of logarithms: $[-i;i]$ and $[s-i;s+i]$.

Our contour looks like

Apart from the cut, there no special points inside our closed contour. Therefore, we can squeeze the contour, pulling it over the cut. Due to our choice the main logarithm branch is at $0<z<s$; when going around $z=i$ to the left bank of the cut $\log\frac{z-i}{z+i}\to \log\frac{z-i}{z+i}+2\pi i$.

So, the integral around the cut $$I(s)=-\frac{1}{8\pi i}\int_{-i}^i\log\frac{z-i}{z+i}\log\frac{z-s+i}{z-s-i}dz-\frac{1}{8\pi i}\int_{i}^{-i}\Big(2\pi i+\log\frac{z-i}{z+i}\Big)\log\frac{z-s+i}{z-s-i}dz$$ $$=-\frac{1}{4}\int_{i}^{-i}\log\frac{z-s+i}{z-s-i}dz=-\frac{1}{4}\Big(s\log\frac{2i-s}{-s}+s\log\frac{-2i-s}{-s}+2i\log\frac{-2i-s}{2i-s}\Big)$$ $$=\frac{s}{2}\log s-\frac{s}{4}\log(s^2+4)-\frac{i}{2}\log\frac{1+s/2}{-i+s/2}$$

Considering the boundary conditions (at $s\to0$ and $s\to\infty$) we have to choose the right branch of logarithm and identify the third term as $\,\,\frac{\pi}{2}-\tan^{-1}(\frac{s}{2})$.

Finally, $$I(s)=\frac{s}{2}\,\log s-\frac{s}{4}\,\log(s^2+4)+\frac{\pi}{2}-\tan^{-1}\Big(\frac{s}{2}\Big)$$