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I am asked to show that $$\lim_{t \to \infty} \int_\frac{1}{t}^t \frac{\sin(ax)}{x}dx $$ converges for all real numbers $a$ and that the value of the converged integral is the same for all $a>0$.

For the first part, I want to apply a previously proven result that if $f$ is continuous and $g'$ is locally integrable on [a,b) with $\lim_{x \to b^- }g(x) = 0$ and $F(t)=\int_a^tf$ is bounded on $[a,b)$, and $g'$ has a constant sign then $\int_a^bfg$ converges. Can I use this theorem here? If we take $g(x)=\frac{1}{x}$ and $f(x)=\sin(ax)$ then clearly $F(t)$ is bounded and $g'$ has a constant sign and $g(x)$ tends to 0. But the condition of local integrability is throwing me off.

If I cannot use this theorem. How may I alternatively prove that the above integral converges? additionally, the proof that the value of the integral is the same for all positive $a$ is eluding me. Any help is greatly appreciated.

Olivier Oloa
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sequenceDerivative
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2 Answers2

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We assume $a>0$, $t>0$.

Hint. One may first perform an integration by parts, $$ \int_{1/t}^t \frac{\sin(ax)}{x}dx =\left[\frac1{x}\cdot\frac{1- \cos(ax)}{a} \right]_{1/t}^t+\int_{1/t}^t \frac{1- \cos(ax)}{ax^2}dx $$ giving $$ \lim_{t \to \infty}\int_{1/t}^t \frac{\sin(ax)}{x}dx =\lim_{t \to \infty}\int_{1/t}^t \frac{1- \cos(ax)}{ax^2}dx\tag1 $$ the latter integrand may be extended as a continuous function over $[0,b]$ ($b\ge1$) satisfying $$ \left| \frac{1- \cos(ax)}{ax^2}\right| \le \frac{2}{ax^2}, \qquad x\ge1, $$ the latter dominating function being integrable over $[1,\infty)$, one deduces the convergence of the initial integral.

From $(1)$, using a standard result one has

$$ \lim_{t \to \infty}\int_{1/t}^t \frac{\sin(ax)}{x}dx =\int_0^\infty \frac{1- \cos(ax)}{ax^2}dx=\frac{\pi}{2} ,\quad a>0. $$

Olivier Oloa
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  • What is the standard result you use in the last point? – sequenceDerivative Nov 10 '16 at 00:03
  • @sequenceDerivative One has $\lim_{t \to \infty}\int_{1/t}^t \frac{\sin(ax)}{x}dx=\int_0^\infty \frac{\sin(ax)}{x}dx$, then with the latter integral make the change of variable $t=ax$, giving $dx/x=dt/t$ and $\int_0^\infty \frac{\sin(ax)}{x}dx=\int_0^\infty \frac{\sin(t)}{t}dt$, then please have a look here: https://en.wikipedia.org/wiki/Dirichlet_integral – Olivier Oloa Nov 10 '16 at 13:23
  • Can you elaborate on the first equality? How do you arrive at $\frac{1-cos(ax)}{a}$ by integrating $sin(ax)$? – Stefan Rickli Aug 06 '20 at 21:45
  • I've just used $\partial_x \left[\cos (ax)\right]=-a \sin (ax)$ giving $$\partial_x \left[\frac{1}{a} - \frac{1}{a}\cos (ax)\right]=\partial_x \left[\frac{1}{a}\right] + \partial_x \left[\frac{1}{-a}\cos (ax)\right]=0+\sin (ax).$$ – Olivier Oloa Aug 07 '20 at 14:06
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Another way is to note that, for $a>0$, we have $$I\left(a\right)=\int_{0}^{\infty}\frac{\sin\left(ax\right)}{x}dx=\frac{1}{2i}\int_{0}^{\infty}\frac{e^{iax}-e^{-iax}}{x}dx $$ and now we can use the Frullani theorem for complex parameters and get $$I\left(a\right)=\frac{\log\left(\frac{ia}{-ia}\right)}{2i}=\frac{\log\left(-1\right)}{2i}=\color{red}{\frac{\pi}{2}}.$$

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