Doubts concerning to an application of Frullani's theorem to $f_k(x)=\frac{2^{-k^2x}-2^{-(k^2+1)x}}{x}$, and Lebesgue convergence theorems

By application of Frullani's theorem for $$a_n=n^2+1$$, $$b_n=n^2$$ where $$n\geq 2$$ and $$f(x)=2^{-x}$$ then RHS in Frullani's integral is obtained for $$n\geq 2$$ as $$\log(1-\frac{1}{n^2}),$$ thus I asked to me (perhaps is easy or my some of the doubts makes no sense, but I want refresh the Lebesgue theorems and know what are my mistakes in myself mathematical reasonings)

Question 1. a) Can then I claim that the integral in LHS converges for each $$n\geq 2$$? (it is precisely one of the hypothesis of the theorem).

b)It is possible give a justification of $$\int_0^\infty 0dx=0$$ (makes sense this expression?) using Lebesgue Monotone Convergence theorem for the nonnegative measurable functions $$f_k=\frac{2^{-k^2x}-2^{-(k^2+1)x}}{x},$$ for $$k\geq 1$$ and previous identity deduce from Frullani's theorem? If the identity deduced is obvious from Lebesgue integral say me, but I would like to know if my application of this convergence theorem was right.

I am apologize for several questions, but I prefer this time that the post was selfcontained. You can give the more concise answer, and if one of my claims is obvious you can say it then I can to say goodbye to these doubts (always the details are welcome).

Question 2. Since we know that $$\log\prod_{n=2}^\infty \left( 1-\frac{1}{n^2} \right)=\log \frac{1}{2}$$ then we should have convergence for $$\sum_{n=2}^{\infty}\int_0^\infty\frac{2^{-k^2x}-2^{-(k^2+1)x}}{x}dx?$$ It is possible does swap the signs $$\sum_{n=2}^\infty$$ and $$\int_0^\infty$$? How is it justified? Thanks in advance.

• My code for Wolfram Alpha were respectively sum 2^(-(n^2+1)x)-2^(-(n^2)x) from n=2 to infinite (next) integrate 2^(-(n^2+1)x)/x from x=0 to infinite (next) integrate (-2^(-2 x-1) (2^x-1) (-2^x+2^x ϑ_3(0, 2^(-x))-2))/x dx from x=0 to infinite
– user243301
May 30 '16 at 10:11
• PLease feel free if you detect a mistake in my computation to tell me.
– user243301
May 30 '16 at 10:12

The theorem in question is that when $f'(x)$ is continuous and the integral converges, then $$\int_0^\infty \frac{f(ax) - f(bx)}x dx = \left(f(0) - f(\infty)\right)\log\left(\frac ba \right).$$

When $a = n^2 + 1, b = n^2$ and $f(x) = 2^{-x}$, the RHS reduces to $\log\left(1 - \frac 1{n^2+1}\right)$. Based on your later integrals, you wanted $a_n = n^2, b_n = n^2 + 1$ instead. But in that case, the RHS reduces to $\log\left(1 + \frac 1{n^2}\right)$, so in either case you are off.

Question 1a) There is no "then" to it (i.e., it doesn't follow from what you already proved - or would have proved without the error). But yes, the LHS converges. $$\lim_{x\to 0}\frac{2^{-ax} - 2^{-bx}}x$$ exists (ref. L'Hopital), so $f$ can be extended to continuous function on $[0, \infty)$. So it is only improper at $\infty$. But there the integrand is monotone (whether increasing or decreasing depends on the relative values of $a$ and $b$ - for $b = a + 1, a > 0$, it will be increasing). It is also trapped between $2^{-ax}$ and $-2^{-bx}$ for large $x$. Since both of those functions are integrable on $[0,\infty)$, it follows that the LHS integrand is integrable as well.

Question 1b) it is always possible to prove the trivial using ridiculously high-powered approaches. This is because in the end, you can just drop the high-powered stuff you've been wasting time on and give the trivial proof after all.

Look at the definition of integration (whatever form of integration is your choice). When the integrand is $0$, all of the sums are also $0$, so their limits are $0$, and so the integral is also $0$. Always. It doesn't matter what the range of integration is. You can even make a tolerable claim that $\int_E 0 = 0$ when $E$ is an unmeasurable set - although for any other integrand on $E$, the integral would be undefined.

2) The Dominated Convergence theorem can be used to justify exchanging the summation and integration. You just need to find a function $g(x)$ such that $$\sum_{n=0}^\infty \frac{2^{-k^2x}-2^{-(k^2+1)x}}{x} \le g(x)$$ and $$\int_0^\infty g(x) dx < \infty$$ To find such a $g$, consider $2^{-(k-1)^2}$.

• Very thanks much for emphatize each of your claims, versus my mistakes. I've studied mathematics, but now I am stuck with some aspects of my reasoning, definitions, theorems,... but I don't want forget the few that I know currently. I've read it, and I have it as reference now.
– user243301
May 30 '16 at 16:25