How should I get the relations eq 3.36 into eq 3.37 in griffiths?

Question

enter image description here How to get this relations...?? Please give me answers.

Answer 1

After checking the reference, I'm quite certain that the formula was intended to be verified for $x>0$. Let me assume this.

Set $r=e^{-\pi x/a}, \theta=\pi y/a, z=re^{i\theta}$ and notice that

$$\sum_{n=1,3 , 5\ldots}\frac{1}{n}e^{-n\pi x/a}sin(n\pi y/a)=\frac{f(z)-f(\bar{z})}{2i},$$ where $f(z)=\sum_{n:odd}\frac{1}{n}e^{-n\pi x/a}e^{in\pi y/a}=\sum_{n:odd}\frac{1}{n}z^n.$

Since $r<1$, $f'(z)=1+z^2+z^4+\cdots$ well converges to $\frac{1}{1-z^2}=\frac{1}{2}(\frac{1}{1-z}+\frac{1}{1+z}).$ After integration we get

$$f(z)=\frac{1}{2}log\frac{1+z}{1-z}.$$

Though we should be careful about in terms of which branch of the complex logarithmic function is $f$ defined, I will proceed without sophistication and concentrate on obtaining the desired formula.

Set $1-r^2+i2rsin\theta=Re^{i\Psi}$, so that $f(z)-f(\bar{z})=\frac{1}{2}log\frac{(1+z)(1-\bar{z})}{(1-z)(1+\bar{z})}=\frac{1}{2}log\frac{1-r^2+i2r sin\theta}{1-r^2-i2r sin\theta}=\frac{1}{2}log\frac{Re^{i\Psi}}{Re^{-i\Psi}}=\frac{1}{2}log(e^{2i\Psi})=\frac{1}{2}2i\Psi$.

We want to show that $$\frac{f(z)-f(\bar{z})}{2i}=\frac{1}{2}tan^{-1}\biggl(\frac{sin(\pi y/a)}{sinh(\pi x/a)}\biggr)=\frac{1}{2}tan^{-1}\bigg(\frac{2sin\theta}{r^{-1}-r}\bigg),$$ i.e.$$tan\bigg(\frac{f(z)-f(\bar{z})}{i}\bigg)=tan\Psi=\frac{2sin\theta}{r^{-1}-r}.$$

This is indeed the case, since by definition $tan\Psi=\frac{Im(Re^{i\Psi})}{Re(Re^{i\Psi})}=\frac{2rsin\theta}{1-r^2}.$

Answer 2

Let $$V(x,y)=\frac{4V_0}{\pi} S.$$ then Use the series: $$\sum_{k=1}^{\infty} \frac{Z^{2k-1}}{2k-1}=\tanh^{-1} Z.$$ Introducing $z=\pi(x+iiy)/a$, let us write The sum as $$S=\sum_{k=1}^{\infty}\left( \frac{e^{-(2k-1)z}-e^{-(2k-1)}\bar z}{2i(2k-1)} \right)=\frac{1}{2i}[\tanh^{-1} e^z-\tanh^{-1} e^{-\bar z}]$$ $$ =\frac{1}{2i}\tanh^{-1} \frac{e^{-z}-e^{-\bar z}}{1- e^{-(z+\bar z)}}= \frac{1}{2i}\tanh^{-1} \frac{e^{-\pi x/a}~ 2i \sin (i\pi y/a)}{1-e^{-2\pi x/a}}$$ $$\frac{1}{2i}\tanh^{-1} \frac{ 2i \sin (\pi y/a)}{e^{\pi x/a}-e^{-\pi x/a}}=\frac{1}{2i} \tanh^{-1} \frac{i\sin \pi y/a}{\sinh \pi x/a}$$ Note that $-i\tanh^{-1}{ig}=\tan^{-1}{g}$, we get $$S=\frac{1}{2} \tan^{-1}\frac{\sin \pi y/a}{\sinh \pi/a}$$ $$\implies V(x,y)=\frac{2V_0}{2} \tan^{-1} \frac{\sin \pi y/a}{\sinh \pi x/a}.$$