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We have the following test prep question, for a measure theory course:

$\forall s\geq 0$, define $$F(s)=\int_0^\infty \frac{\sin(x)}{x}e^{-sx}\ dx.$$

a) Show that, for $s>0$, $F$ is differentiable and find explicitly its derivative.

b) Keeping in mind that $$F(s)=\int_0^\pi \frac{\sin(x)}{x}e^{-sx}\ \ dx\ +\int_\pi^\infty \frac{\sin(x)}{x}e^{-sx}\ dx,$$ and conveniently doing integration by parts on the second integral on the right hand side of the previous equation, show that $F(s)$ is continuous at $s=0$. Calculate $F(s)\ (s\geq 0)$.

Since it's a measure theory course, I'm thinking there are methods involving the things you typically learn in these courses, and I think Lebesgue's Dominated Convergence Theorem will play a role, because I was looking at books by Bartle and Apostol, and they both have similar exercises or theorems, and both use LDCT.

Also, I suppose these proofs regarding continuity or differentiability could be done with standard calculus stuff (like $\epsilon$'s and $\delta$'s or the actual definition of a derivative), but I want to avoid these methods and focus on what I should be learning from the class.

I think I have part (a), or at least a good idea, based on the Bartle book. If I let $f(x,s)=\frac{\sin(x)}{x}e^{-sx}$, I just need to find an integrable function $g$ such that $\big|\frac{\partial f}{\partial s}\big|\leq g(x)$ (after showing that partial does exist, of course :) ). And then, $$\frac d{ds}F(s)=\int _{\mathbb{R}^+}\frac{\partial f}{\partial s}\ dx.$$ Please correct me if I'm mistaken, or missing something.

Now, for part (b) I'm a little stumped. In the Apostol book, the case $s>0$ is done explicitly, but I read through it and it didn't help me. Looking at the Bartle book, I get the idea of defining $f_n=(x,s_n)$, where $s_n=\frac1{n+1}$ or some such sequence that goes to zero. Then, somehow, maybe, LDCT kicks in (but I guess I'd have to find a function what would dominate these $f_n$). I also don't really see the point in dividing the integral into the two parts up there, so I must be missing something.

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2 Answers 2

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For part a), your idea is good: take $g(x):=e^{-x}\chi_{(1,+\infty)}+\chi_{[0,1]}$, which is integrable.

For part b), the first term converges as $s\to 0$ to $\int_0^\pi\frac{\sin t}tdt$, by LDTC. For the second one, we write \begin{align} \int_{\pi}^{+\infty}\frac{\sin t}te^{-st}dt&=\int_\pi^{+\infty}\sin t\frac{e^{-st}}tdt\\ &=\left[-\cos t\frac{e^{-st}}t\right]_{t=\pi}^{t=+\infty}+\int_\pi^{+\infty}\cos t\left(-s\frac{e^{-st}}t-\frac{e^{-st}}{t^2}\right)dt\\ &=\frac{e^{-s\pi}}\pi-\int_{\pi}^{+\infty}\frac{\cos t}{t^2}e^{-st}dt-s\int_\pi^{+\infty}\cos t\frac{e^{—st}}tdt. \end{align} By LDTC, the first two terms converge to $\frac 1{\pi}-\int_{\pi}^{+\infty}\frac{\cos t}{t^2}dt$, and integrating by parts we notice it's equal to $\int_\pi^{+\infty}\frac{\sin t}tdt$.

So we have to show that $\lim_{s\to 0}s\int_\pi^{+\infty}\cos t\frac{e^{—st}}tdt=0$. To see that, we integrate by parts.

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You're correct that LDCT is the right theorem for this problem.

For part A, try explicitly writing out the definition of the derivative for this function and using LDCT to interchange the limit in the definition of the derivative with the integral- this isn't quite the same as bounding the derivative of f by an integrable function. Together with the linearity of integration, this will let you move the derivative "inside" the integral as you wanted to do.

For part B, you want to use LDCT on both of the integrals on the right hand side to take a limit as s goes to 0, but you run into the problem that LDCT doesn't actually apply to the second integral! The integral is split up so that the product term arising from the integration by parts will cancel out, and you should be able to apply LDCT to the new integral you get. You can then use the continuity of the integrand to finish the problem.

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