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Finite Strip

Full disclosure: This is technically a "homework" question but not really. What I mean is that my professor gave us free reign to use Mathematica to simply get the answer to the integral but I am trying to go the extra mile and understand the full derivation, but I'm stuck.

The problem is solving the Laplace equation for a finite strip. The solution which I've already derived is:

$$V(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\sin{k}}{k}e^{ikx}e^{-|k|y}dk\tag{1}$$

He asks us to try to solve this integral by hand (and if we can't, to feel free to simply use Mathematica to get the answer) which I've been working on for a couple of hours. I've gotten as far as changing the integral to:

$$V(x,y)=2\int_{0}^{\infty}dk\frac{\sin{k}}{k}e^{ikz}-2\int_{0}^{-\infty}dk\frac{\sin{k}}{k}e^{ikz^*}\tag{2}$$

Where $z\equiv(x+iy)$. This is something like the Fourier transform that takes $\text{sinc}(k)\rightarrow\tilde{\text{sinc}}(z)$ but I've never evaluated a Fourier transform in the complex plane. I know I'm on the right track as the correct answer is:

$$V(x,y)=-\frac{2}{\pi}\Im(\text{arctanh}(z^{-1}))\tag{3}$$

But I'm nevertheless stuck and would appreciate a hint or a nudge in the correct direction so that I see how to proceed.

I have an inkling that the answer involves expressing $z^{-1}$ as $z^{-1}=\frac{d}{dz}\ln{z}$ and then evaluating the integral until you get something like $\frac{1}{z^*}-\frac{1}{z}$ and expressing that as $\frac{d}{dz}(\ln(z^*)-\ln(z))=\frac{d}{dz}\ln{\frac{z^*}{z}}$ which is closely related to arctan functions as discussed here but I don't quite see how to get there.

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Commented Nov 16, 2022 at 17:30

2 Answers 2

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Too long for a comment

Using $\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$ and noting, that only $\cos kx$ from $e^{ikx}=\cos kx+i\sin kx\,$ survives integration (the even part of the integrand), we can present the integral in the form $$V(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\sin{k}}{k}e^{ikx}e^{-|k|y}dk=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\sin{k}\cos kx}{k}e^{-|k|y}dk$$ $$=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\sin{k(x+1)-\sin k(x-1)}}{k}e^{-|k|y}dk=\frac{1}{\pi}\Im\int_0^\infty\frac{e^{-k\big(y-i(x+1)\big)}-e^{-k\big(y-i(x-1)\big)}}{k}dk$$ This is the type of Frullani integral which is, strongly speaking, applicable to a real function, but at some additional conditions can be used for a complex function as well (the discussion about the similar case, for example, here).

Supposing that these conditions are met and denoting $z=x+iy$, $$V(x,y)=\frac{1}{\pi}\Im\ln\frac{y-i(x-1)}{y-i(x+1)}=\frac{1}{\pi}\Im\ln\frac{x+iy-1}{x+iy+1}=\frac{1}{\pi}\Im\ln\frac{1-\frac{1}{z}}{1+\frac{1}{z}}=-\frac{2}{\pi}\Im\operatorname{arctanh}\Big(\frac{1}{z}\Big)$$

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Let $V(x,y)=\frac1{2\pi}\int_{-\infty}^\infty \frac{2\sin(k)}{k}e^{-|k|y}e^{ikx}\,dk$ for $y\ge 0$. Then, we can write

$$\begin{align} V(x,y)=\frac2\pi \int_0^\infty \frac{\sin(k)\cos(kx)}{k}e^{-ky}\,dk\tag1 \end{align}$$

We use Feynman's trick to evaluate the integral in $(1)$. Differentiating $(1)$ with respect to $y$ (valid due to the uniform convergence of the integral of the differentiated integrand for $y$ bounded away from $0$) reveals

$$\begin{align} \frac{\partial V(x,y)}{\partial y}&=-\frac2\pi \int_0^\pi \sin(k)\cos(kx)e^{-ky}\,dy\\\\ &=\frac1\pi \left(\frac{(x-1)}{(x-1)^2+y^2}-\frac{(x+1)}{(x+1)^2+y^2}\right)\tag2 \end{align}$$

Integrating with respect to $y$, we find

$$\begin{align} V(x,y)&=\frac1\pi \left(\arctan\left(\frac{y}{x-1}\right)-\arctan\left(\frac{y}{x+1}\right) \right)\\\\ &=\frac1\pi \left(\arctan\left(\frac{x+1}{y}\right)-\arctan\left(\frac{x-1}{y}\right)\right) \end{align}$$

And we are done!

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  • $\begingroup$ @codypayne Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Commented Nov 16, 2022 at 17:30

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