# Prove that $\int_a^\infty f(x)\sin(e^x) \, dx$ conditionally converges.

Let $$f$$ be a bounded and with a continuous derivative at the interval $$[a,\infty)$$.

The integral: $$\displaystyle \int_a^\infty f(x) \, dx$$ diverges.

Also: $$\exists t> a, \forall x>t: f'(x) < f(x)$$

Prove that the $$\displaystyle \int_a^\infty f(x) \sin(e^x) \, dx$$ conditionally converges.

## What i tried:

So i want to show that it diverges in its absolute value and converges in its "normal" value.

For diverges in its absolute value

We know that: $$|\sin(e^x)| < 1$$

Therefore we can write: $$\int_a^\infty |f(x)| \leq \int_a^\infty |f(x)\sin(e^x)| \, dx$$

But we know that the left integral diverges from the question, therefore, by comparison test: $$\int_a^\infty |f(x)\sin(e^x)| \, dx$$ diverges.

Now the problem in proving converges for the "normal" function.

I thought to use Dirichlet test, but i dont see how to say that $$f(x)$$ is decreasing monotonic or to talk about the limit.

I must say it very sounds like drichlet test, but i cant see how it feets...

Couldnt think of other functions for dirichle

So i thought about the comparison test, yet couldn't think of a converging function that will fit.

In the end, i am stuck.

Those are my homework, so i prefer a hint than a solution.

Thank you.

• It's the opposite, $|f(x)\sin(e^x)|\leq |f(x)|$, so the comparison test will not work so easy.
– Mark
Apr 6, 2020 at 18:19
• See my answer below. Apr 6, 2020 at 19:00
• @Alon Did you ever find a better solution for this problem?
– Yly
Sep 1, 2020 at 5:43

I suggest to write $$f(x) \sin(e^x) = e^{-x} f(x) e^{x} \sin(e^x)$$ and to integrate by parts considering $$u(x) = e^{-x} f(x)$$ and $$v'(x) = e^{x} \sin(e^x).$$ $$\int_a^\infty f(x) \sin(e^x)\, dx = -\left[e^{-x}f(x)\cos(e^x)\right]_a^\infty + \int_a^\infty e^{-x}(f'(x) -f(x)) \cos(e^x)\, dx.$$

Using your assumption (I guess it is $$|f'| \leq |f|$$) together with the boundness of $$f$$, you can prove that the integral $$\int_a^\infty e^{-x}(f'(x) -f(x)) \cos(e^x)\, dx$$ is convergent.

• Thank you, but the condition $f' < f$ is as i wrote and without absolute value
– Alon
Apr 6, 2020 at 22:11
• Oh i think i understand. you mean that $(f'(x)-f(x)) = L, L \in R$ as its bounded, the same idea goes for $|cos(e^x)| \leq 1$ so call it $P$ and we know that $\lim_{x \to \infty}e^{-x} = 0$ Therefore we have that limit that goes to something like: $0 \cdot L \cdot P = 0$. Therefore our integral has a finite limit, therefore its converges.
– Alon
Apr 7, 2020 at 1:52
• Do you have an idea (or someone have) for the case with the absolute value? What i wrote was wrong, as Mark explained in the comments
– Alon
Apr 7, 2020 at 2:17
• Having the limit zero at infinity doesn't imply necessarily that the function is integrable. Take for example $f(x) = 1/x$ which is not integrable on $[1, \infty[.$ For the absolute value, if you assume that $|f'|\leq |f| \leq M,$ then you know that $|e^{-x}(f'(x) - f(x))\cos(e^x)| \leq 2M e^{-x}$ which is integrable on $[a, \infty[$ Apr 7, 2020 at 13:19
• The assumption $f' < f$ doesn't imply that $f'$ is bounded. Apr 7, 2020 at 13:27

With your hypothesis $$f' (not $$|f'|<|f|$$) this problem seems very tricky. Here's a solution sketch I've been able to work out. To show that the integral converges:

• First note that the convergence only depends on the behavior of $$f$$ at large $$x$$, so WLOG we can assume that $$f' everywhere.
• Then use the integration by parts and integration factor of @A.Pi's answer.
• Note that boundedness of $$f$$ implies that $$\int_a^\infty e^{-x}f(x)\cos(e^x)dx$$ converges absolutely, so we need only show that the remaining term $$\int_a^\infty e^{-x}f'(x)\cos(e^x)dx$$ converges.
• Define $$f'_+(x):= \max\{f'(x),0\}$$ and $$f'_-(x):=\max\{-f'(x),0\}$$, so that $$f' = f'_+ - f'_-$$. Then we have $$|f'(x)| = f'_+ + f'_-$$ and thus $$\left|\int_a^b e^{-x} f'(x) \cos(e^x) dx\right| \leq \int_a^b e^{-x} \left(f'_+ + f'_-\right) dx$$.
• By boundedness of $$f$$, there is some $$M$$ such that $$|f|\leq M$$, and thus on any interval $$[b,c]$$ we have $$2M\geq |f(c)-f(b)| = \left|\int_b^c f' dx\right| = \left|\int_b^c f'_+ dx - \int_b^c f'_- dx\right|$$ Furthermore, since $$f' the term $$\left|\int_b^c f'_+ dx\right|\leq M(c-b)$$, and thus $$\int_b^c f'_- dx\leq 2M + M(c-b)$$. This is what we will use to "bound $$f'$$ from below", which is what makes this problem hard.
• On an interval $$[a,b]$$, to show that $$\int_a^b e^{-x} (f'_+ + f'_-) dx$$ converges as $$b\rightarrow \infty$$, chop up $$[a,b]$$ into subintervals $$[a,a+1]$$, $$[a+1,a+2]$$, $$[a+2,a+3]$$, $$\dots$$, $$[a+n,b]$$, where $$n=\lfloor b-a \rfloor$$. The integral over $$[a+k,a+k+1]$$ or over $$[a+n,b]$$ is at most $$4Me^{-a-k}$$ by the bound above on $$\int f'_+ dx$$ and $$\int f'_- dx$$. Thus the sum over these intervals converges geometrically as $$k\rightarrow \infty$$. This proves the convergence we wanted to show.

To show that the integral $$\int_a^\infty \left|f(x) \sin(e^x)\right| dx$$ does not converge, first note that the condition $$f' implies that $$f>0$$, because otherwise $$f$$ would diverge to $$-\infty$$, contradicting its boundedness. Hence the above integral is $$\int f(x) |\sin(e^x)| dx$$.

Now use the same $$e^{\pm x}$$ integration trick as before, assume WLOG that $$e^a=2\pi k$$ for some integer $$k$$ (otherwise increase $$a$$ a little bit to make this true), and note that

$$\int_a^x e^y |\sin(e^y)| dy = 2n(x)+1 - (-1)^{n(x)}\cos(e^x)$$

where $$n(x) = \lfloor \frac{e^x-e^a}{\pi}\rfloor$$ counts how many times $$\sin(e^x)$$ switches sign. Then following the same analysis as above, we find that

$$\int_a^b |f(x) \sin(e^x)| dx = \left[e^{-x} f(x) \left(2n(x)+1 - (-1)^{n(x)}\cos(e^x)\right)\right]_a^b - \int_a^b e^{-x} (f'(x)-f(x)) \left(2n(x)+1 - (-1)^{n(x)}\cos(e^x)\right)dx$$ As $$b\rightarrow \infty$$, everything in the first term vanishes except $$2n(b)e^{-b}f(b)$$, which $$\rightarrow \frac{2}{\pi}f(b)$$. In the remaining integral, by similar arguments as above, everything converges absolutely except $$\int_a^b 2n(x)e^{-x}(f'(x)-f(x)) dx$$. Note that $$n(x) = \frac{e^x}{\pi} + \epsilon(x)$$, where $$\epsilon(x)$$ is bounded. Thus the preceding integral can be written as $$\int_a^b \frac{2}{\pi} (f'(x)-f(x)) dx$$ plus an absolutely convergent term. $$\int_a^b \frac{2}{\pi} f'(x) dx$$ cancels with the $$\frac{2}{\pi} f(b)$$ term above, and we are left with $$\frac{2}{\pi}\int_a^b f(x) dx$$, which diverges by hypothesis.

This proves that $$\int_a^\infty |f(x) \sin(e^x)| dx$$ diverges.

• I talked to the tutor from the university. He said, dont look for tricks, its a very standard question.... use dirichle...
– Alon
Apr 8, 2020 at 21:11
• @Alon Here's what Wikipedia says about that: en.wikipedia.org/wiki/Dirichlet%27s_test#Improper_integrals The problem is that none of the functions in this question are monotonically decreasing.
– Yly
Apr 8, 2020 at 21:23
• @Alon Are you sure you have the hypotheses on your question correct? If instead of $f'<f$ you had something along the lines of $f'<0$, then you could apply Dirichlet.
– Yly
Apr 8, 2020 at 21:40
• Yea, i know dirichle, and the hypothesis correct, i also wanted it to be under zero therefore it would be much easier to match to dirichle.
– Alon
Apr 8, 2020 at 23:12
• @Alon I posit that your tutor is wrong about this being a standard question, but perhaps I and everyone else who's responded to your questions are just missing something. At any rate, I believe my solution is correct. If you or your tutor devise a simpler one, please post it as an answer to your question.
– Yly
Apr 9, 2020 at 18:51