TITLE: Why is it true that $\int_0^\infty \sin(qx) dx = 1/q$.
QUESTION [6 upvotes]: In a Quantum Mechanics class the professor pointed out a calculation from Zetilli's textbook on Quantum Mechanics (page 630).
There he says that:
$$\int_0^\infty \sin(qx) dx = \lim_{\lambda \to 0} \int_0^\infty e^{-\lambda x} \sin(qx) dx$$
Then he writes the $\sin$ as complex exponentials and reaches the result that the integral is $1/q$. WHAT? I was taught at my calculus course that this integral does not converge. For me this result is plain wrong, and I do not understand it at all. I even don't see why the limit can be imposed like that. Any thoughts on this "trick"?

REPLY [3 votes]: Since
$$
\newcommand{\Im}{\operatorname{Im}}
\int_0^\infty\sin(qx)\,\mathrm{d}x=
\Im\left(\int_0^\infty e^{iqx}\,\mathrm{d}x\right)\tag1
$$
in the spirit of analytic continuation, we can look at
$$
f(q)=\int_0^\infty e^{iqx}\,\mathrm{d}x\tag2
$$
which is well-defined for $\Im(q)\gt0$, and see how $f$ behaves as $\Im(q)\to0$.
When $\Im(q)\gt0$, we get, via contour integration,
$$
\begin{align}
f(q)
&=\frac iq\int_0^{-iq\infty} e^{-x}\,\mathrm{d}x\tag{3a}\\
&=\frac iq\int_0^\infty e^{-x}\,\mathrm{d}x\tag{3b}\\
&=\frac iq\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: substitute $x\mapsto\frac{ix}q$
$\text{(3b)}$: integration over $\gamma=[0,R]\cup[R,-iqR]\cup[-iqR,0]$ is $0$
$\text{(3c)}$: integrate

Then, by analytic continuation, as $\Im(q)\to0$, $\frac iq$ behaves nicely. Therefore, we say for real $q$:
$$
\int_0^\infty(\cos(qx)+i\sin(qx))\,\mathrm{d}x=\frac iq\tag4
$$
which says not only that $\int_0^\infty\sin(qx)\,\mathrm{d}x=\frac1q$, but also that $\int_0^\infty\cos(qx)\,\mathrm{d}x=0$ (even though neither of these integrals converges in the classical sense).

Let's verify the integral of $\cos(qx)$:
$$
\begin{align}
\int_0^\infty\cos(qx)\,\mathrm{d}x
&=\int_0^\infty\sin(\pi/2+qx)\,\mathrm{d}x\tag{5a}\\[6pt]
&=\int_{\frac\pi{2q}}^\infty\sin(qx)\,\mathrm{d}x\tag{5b}\\
&=\color{#C00}{\int_0^\infty\sin(qx)\,\mathrm{d}x}\color{#090}{-\int_0^{\frac\pi{2q}}\sin(qx)\,\mathrm{d}x}\tag{5c}\\[3pt]
&=\color{#C00}{\frac1q}\color{#090}{+\left.\frac1q\cos(qx)\,\right|_0^{\frac\pi{2q}}}\tag{5d}\\[9pt]
&=0\tag{5e}
\end{align}
$$
Explanation:
$\text{(5a)}$: $\cos(x)=\sin(\pi/2+x)$
$\text{(5b)}$: substitute $x\mapsto x-\frac\pi{2q}$
$\text{(5c)}$: difference of integrals
$\text{(5d)}$: integrate
$\text{(5e)}$: evaluate