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I recently came across the fact that $\displaystyle\int_{\frac{m}{a}}^{\frac{n}{a}}\frac{f(ax)}{x}\,dx$ is independent of the value of $a$. This lead me to this integral here: $\displaystyle \int_0^\infty \frac{\sin(ax) \cos(bx)}{x}\,dx$.

From the above result of independence, it is clear that:

$ \begin{align} \int_0^\infty \frac{\sin(bx+ax)}{x}\,dx &= \int_0^\infty \frac{\sin(bx-ax)}{x}\,dx\\ \int_0^\infty \frac{\sin(bx+ax)}{x}\,dx - \int_0^\infty \frac{\sin(bx-ax)}{x}\,dx &= 0\\ \int_0^\infty \frac{\sin(bx+ax) - \sin(bx-ax)}{x}\,dx&= 0\\ 2\int_0^\infty \frac{\sin(ax) \cos(bx)}{x}\,dx&= 0\\ \int_0^\infty \frac{\sin(ax) \cos(bx)}{x}\,dx&= 0. \end{align} $

However, it's obvious that this is incorrect by counterexample, say $a = 2$ and $b = 1$. Where is the mistake in this proof? Is there a restriction on $a$ and $b$ that I need to take into consideration? Any help or guidance would be greatly appreciated!

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    $\begingroup$ What is your $f$ when you apply the independence? $\endgroup$ – Ramanujan Jun 21 at 1:52
  • $\begingroup$ $f(x)$ is $\sin(x)$. For the LHS integral, the constant is $b+a$, for the RHS integral the constant is $b-a$. $\endgroup$ – LogicAndTruth Jun 21 at 1:53
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For finite $m$ and $n$, it is true that the value of

$$ \int_{m/a}^{n/a} \frac{\sin(ax)}{x}\, \mathrm dx $$

is independent of $x.$ But when you integrate to $\infty,$ you need to watch the sign. In particular, if $p$ and $q$ have opposite signs, then

$$ \lim_{n \to \infty} \frac np = - \lim_{n \to \infty} \frac nq,$$

and therefore

$$ \int_0^\infty \frac{\sin(px)}{x}\, \mathrm dx = \int_0^{-\infty} \frac{\sin(qx)}{x}\, \mathrm dx = - \int_0^\infty \frac{\sin(qx)}{x}\, \mathrm dx. $$

Set $px = bx + ax$ and $qx = bx - ax$ in your calculations; then for $a = 2$ and $b = 1,$ you have $p = 3$ and $q = -1,$ therefore

\begin{align} \int_0^\infty \frac{\sin(2x) \cos(x)}{x}\, \mathrm dx &= \frac12\left(\int_0^\infty \frac{\sin(3x)}{x}\, \mathrm dx - \int_0^\infty \frac{\sin(-x)}{x}\, \mathrm dx\right) \\ &= \frac12\left(\int_0^\infty \frac{\sin(3x)}{x}\, \mathrm dx + \int_0^\infty \frac{\sin(3x)}{x}\, \mathrm dx\right) \\ &= \int_0^\infty \frac{\sin(3x)}{x}\, \mathrm dx \\ &= \int_0^\infty \frac{\sin(x)}{x}\, \mathrm dx. \end{align} Neither side is zero.

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  • $\begingroup$ Great explanation, thank you for this! So the restriction will be $b > a$ in this case $\endgroup$ – LogicAndTruth Jun 21 at 2:27
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    $\begingroup$ Yes, with that restriction the original calculation works. $\endgroup$ – David K Jun 21 at 2:28

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