$\int_0^\infty \frac{\sin x}{x}dx$ using Feynman's technique While watching this video, he says that
$$\lim_{b\to\infty}\int_0^\infty \frac{\sin x}{x}e^{-bx}dx\to 0$$
since the integrand tends to $0$. However I think that it must be an indeterminate form, since $\infty\times0$ may not always be $0$
Am I right in saying this? If yes, how else can we evaluate the integration constant?

Reference picture:

 A: As Professor Vector comments, since $\left|\frac{\sin(x)}x\right|\le1$,
$$
\begin{align}
\left|\int_0^\infty\frac{\sin(x)}x\,e^{-bx}\,\mathrm{d}x\right|
&\le\int_0^\infty e^{-bx}\,\mathrm{d}x\tag1\\
&=\frac1b\tag2
\end{align}
$$
Thus, the integral vanishes as $b\to\infty$.
We can then evaluate the integral using Feynman's technique:
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}b}\int_0^\infty\frac{\sin(x)}x\,e^{-bx}\,\mathrm{d}x
&=-\int_0^\infty\sin(x)\,e^{-bx}\,\mathrm{d}x\tag3\\
&=-1+b\int_0^\infty\cos(x)\,e^{-bx}\,\mathrm{d}x\tag4\\
&=-1+b^2\int_0^\infty\sin(x)\,e^{-bx}\,\mathrm{d}x\tag5\\
&=-\frac1{1+b^2}\tag6
\end{align}
$$
Explanation:
$(3)$: differentiate under the integral (Feynman's technique)
$(4)$: integrate by parts
$(5)$: integrate by parts
$(6)$: add $\frac{b^2}{1+b^2}$ times $(1)$ to $\frac1{1+b^2}$ times $(3)$
Using $(2)$, solving $(6)$ gives
$$
\begin{align}
\int_0^\infty\frac{\sin(x)}x\,e^{-bx}\,\mathrm{d}x
&=\frac\pi2-\tan^{-1}(b)\tag{7}\\
&=\tan^{-1}\left(\frac1b\right)\tag{8}
\end{align}
$$
