The origin of this question comes from watching this YouTube video at 8:00.
The equation to evaluate is $$f(x)=\frac{1}{i(j-k)}e^{i(j-k)x}|_{-\pi}^{\pi}$$
However, this equation can be evaluated normally as long as $j\neq k$. For case where $j=k$, the video says to use L'Hospital's Rule and it will evaluate to $x$ for $j=k$.
I tried just that but came out to be $-1/x$.
Here's my attempt: The L'Hospital's Rule says $$\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}$$.
Here I use j instead of x:
$$f(j)=e^{i(j-k)x}=e^{ijx-ikx}$$ and $$g(j)=i(j-k)$$
and therefore $$f'(j)=\frac{1}{ix}e^{i(j-k)x}$$
and $$g'(j)=i$$
$$\lim_{j\to k}\frac{f'(j)}{g'(j)}=\frac{\frac{1}{ix}e^{i(j-k)x}}{i}=\frac{1}{ix}/i=\frac{1}{-x}$$
Where did I do wrong?
EDIT: I mistook derivate with integral for $f'(j)$
I don't agree with him in that you should evaluate the answer to the integral for $j = k$ using some trick. It's way easier to simply evaluate $\langle{\psi_j, \psi_j \rangle}$ directly. We can just plug this into the integral expression and get $$ \langle \psi_j, \psi_j \rangle = \int_{-\pi}^\pi e^{ijx}e^{-ijx} dx = \int_{-\pi}^\pi 1 dx = 2\pi $$