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It is well known that $\int_0^\infty \frac{\sin ax}{x}dx =\frac{\pi}{2}$. Is this true if $a=0$? I don't think so. Can someone confirm this. Thanks.

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The identity comes from $u=ax$, $\dfrac{du}{a}=dx$, so $$\int_0^{\infty}\dfrac{\sin(ax)}{x}dx=\int_0^{\infty} \dfrac{\sin(u)}{u/a}\dfrac{du}{a}=\int_0^{\infty} \dfrac{\sin(u)}{u}du=\frac{\pi}{2}$$

However, implicitly we used $a>0$ when evaluating the bounds of the integral, since we want $u_0=a\cdot 0$ and $u_{\infty}=a\cdot \infty$. If $a=0$, then the substitution doesn't work, and direct substitution gives the integral to be $0$.

Also, if $a<0$, then we can write $a=-a'$, $a'>0$, and pull out the negative sign to get $$-\int_0^{\infty} \dfrac{\sin(a'x)}{x}dx=-\frac{\pi}{2}$$

Moya
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  • $0 \cdot \infty$ is never interpreted as $0$. Integral is zero due to the simple substitution and keeping limits as they are. – Kaster Aug 21 '15 at 02:06
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    While yes I agree you're right (mistake on my part) $0\cdot \infty$ is interpreted as $0$ in some real analysis settings. That's in the first chapter of Real and Complex analysis. – Moya Aug 21 '15 at 02:10