# Why is this integration proof incorrect?

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!

• What is your $f$ when you apply the independence? – Ramanujan Jun 21 at 1:52
• $f(x)$ is $\sin(x)$. For the LHS integral, the constant is $b+a$, for the RHS integral the constant is $b-a$. – LogicAndTruth Jun 21 at 1:53

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.

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