I am trying to figure the value of $$\lim_{A \to \infty} \int_{(0,A)} \frac{\sin(x)}{x} \,dx$$ using Fubini's theorem.

As for now, I am thinking to substitute $1/x = \int_{(0,\infty)} \exp(−nx) \, dn$, thus

$$\lim_{A\to\infty}\int_{(0,A)} \frac{\sin(x)}{x} \, dx = \lim_{A\to\infty}\int_{(0,A)} \bigg[\sin(x) \int_{(0,\infty)}\exp(−nx)dn\bigg] \,dx.$$

From here, using Fubini's theorem, I just need to change the order of integration and I got $\pi/2$.

The thing is, in order to justify the use of Fubini's theorem, either the condition "absolute value of $\sin(x)\exp(-nx) < \exp(-nx) < \infty$ is integrable on $(0,A)\times(0,\infty)$" or "$\sin(x)\exp(-nx)$ is a non negative increasing function" need to be satisfied (correct me if I'm wrong) which is not satisfied.

Can someone help me to justify the use of Fubini's theorem on this particular problem? Thank you


Your second condition is false. The correct condition would read "$\sin(x)\exp(-nx)$ is a non-negative measurable function". There is no need for anything like "increasing" here.

Anyway, you cannot use this condition here because $\sin(x)$ can also attain negative values.

But you should be able to verify that

\begin{eqnarray*} \int_{(0,A)}\int_{(0,\infty)} |\sin(x) \exp(-nx)| \, dn \, dx &=& \int_{(0,A)} |\sin(x)| \cdot \int_{(0,\infty)} e^{-nx} \, dn \, dx \\ &=& \int_{(0,A)} \frac{|\sin(x)|}{x} \, dx <\infty \end{eqnarray*}

is indeed finite, so that the first condition is satisfied.

To see that this is finite, note that $\frac{|\sin(x)|}{x} =\bigg|\frac{\sin(x)}{x}\bigg| \to 1$ as $t \to 0$, so that the integrand is continuous on $[0,A]$.

  • $\begingroup$ So, is it true that in order to use fubini's theorem, the integrand must either be 1) integrable on a certain domain or 2) integrand must be a non negative measurable function ? $\endgroup$ – mohlee Feb 6 '15 at 4:15
  • $\begingroup$ @mohlee: Yes, exactly. The domain should also be measurable, of course. $\endgroup$ – PhoemueX Feb 6 '15 at 6:33

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