Why does FourierTransform[] converge while same integral manually written does not?

Question

I'm studying the quantum mechanics of an infinite square well from a computational standpoint.

My eigenfunctions are defined as $u_n(x)=\sqrt{2/L}\sin\left(n\pi x/L\right),\quad 0 \le x \le L, \quad n=1,2,3,\dots$ and zero outside of the $[0,L]$ interval.

I'd like to transform the wave function from position domain to the momentum domain. My book says that this is achieved by writing the Fourier transform of $u_n(x)$:

$\phi_n(k)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} u_n(x)e^{-i k x}dx$

ClearAll["Global`*"];
L = 1;
u[n_, x_] := Sqrt[2/L] Sin[n π x]
φ[n_, k_] := Assuming[{n ∈ Integers, {k, x} ∈ Reals},
                1/(Sqrt[2π]) Integrate[u[n, x] Exp[-I k x], {x, -∞, ∞}]]

Integrate::idiv: Integral of E^(-I k x) Sin[n π x] does not converge on {-∞,∞}. >>

If I, though, use FourierTransform[], I get the result:

FourierTransform[u[n, x], x, k, FourierParameters -> {0, -1}]

-I Sqrt[π] DiracDelta[k - n π] + I Sqrt[π] DiracDelta[k + n π]

But my book says that the result should be:

$\phi_n(k) = \sqrt{\pi L} n \frac{1-(-1)^n e^{-i L k}}{(n\pi -L k)(n\pi + Lk)}$

I tried to get Mathematica to fully simplify the output of FourierTransform[], but I/it couldn't.

So my questions are:

1. Why does the integral not converge when written manually but it does converge when it is being calculated via FourierTransform[] ?

2. How could I prove that the output of FourierTransform[] is equivalent to the formula that is written in my book for $\phi_n(k)$ ?

Answer 1

Your function u is defined only for 0..L, then simply apply the integration over that range only.

ClearAll["Global`*"];
L = 1;
u = Sqrt[2/L] Sin[n Pi x/L];
mma = Assuming[Element[n, Integers],1/(Sqrt[2 Pi]) Integrate[u  Exp[-I k x], {x, 0, L}]];
book = Sqrt[Pi L] n (1 - (-1)^n Exp[-I L k])/((n Pi - L k) (n Pi + L k ))

Simplify[book - mma]

On the question why sometimes FourierTransform seems to handle things better with some function vs. when applying the Integrate command directly from the book definition of FourierTransform, I think this has to do to some assumptions build in Fourier Transform about the function itself and may be other subtle things, and FourierTransform might apply special algorithms to integrate the function that might not be used by the general purpose Integrate at some point.