# Prove that $\int_{1}^{+\infty} \arctan\left(\frac{1}{t}\right)\tanh(t)\operatorname{sin}(t)dt$ exists and is a finite value.

I'm trying to prove that this limit

$$\int_{1}^{+\infty} \arctan\left(\frac{1}{t}\right)\tanh(t)\operatorname{sin}(t)dt$$

exists and it's finite.

I have proved that

$$\int_{1}^{+\infty} \left|\arctan\left(\frac{1}{t}\right)\tanh(t)\operatorname{sin}(t) \right|dt = +\infty$$ so I can't use summability.

My other attempt to solve the given integral is the fact that $$f(t)=\arctan\left(\frac{1}{t}\right)\tanh(t)\sin(t)$$ have the same behaviour of the function $$g(t)=\frac{\sin(t)}{t}$$ as $$t\to +\infty$$ and I succeded in proving that $$\int_{1}^{+\infty} g(t)dt$$ converges.

But I know from theory that the asymptotic method can only be used if the integrands have a constant sign. So how can I use what I achieved to prove that the given integral converges? And, more in general, when the integrand has a variable sign what criterion can I use? Can I use something similar to the asymptotic method to deduce that two improper integrals have the same behavior?

PS: another attempt was to use the integral criterion for the series with the general term $$f(n)$$ and the Dirichlet's test, but the problem is that I can't use the integral criterion since the integrand is not decreasing. So I have another question: is there a way to extend the integral criterion? (for example for functions that are not decreasing)

• I don't know what sen is. Jul 25, 2021 at 11:14
• Apparently, sen stands for “seno” which is Spanish for “sine”. A simple search reveals this. Jul 25, 2021 at 11:43
• @TymaGaidash With which search engine? I got this when I put sen(x) into my search bar. (OK, I can guess which search engine, but the point is, Gerry's comment is far from unreasonable, especially given the site is in English.) Jul 25, 2021 at 11:51
• If you read this board a lot, you will see "sen" from time to time. Jul 25, 2021 at 12:01
• Sorry i am used to write "sen" in place of "sin" and vice versa. And i don't notice which notation i use. I edited it :D Jul 25, 2021 at 12:24

There exist bounded functions $$f(t)$$ and $$g(t)$$ such that $$\arctan \left( {\frac{1}{t}} \right) = \frac{1}{t} + f(t)\frac{1}{{t^3 }},\qquad \tanh t = 1 + g(t)e^{ - 2t}$$ for $$t>1$$. Thus, \begin{align*} &\int_1^M {\arctan \left( {\frac{1}{t}} \right)\tanh t\sin tdt} \\ & = \int_1^M {\frac{{\sin t}}{t}dt} + \int_1^M {g(t)e^{ - 2t} \frac{{\sin t}}{t}dt} + \int_1^M {f(t)\frac{1}{{t^3 }}\tanh t\sin tdt} \end{align*} for any $$M>0$$. Here $$\left| {\int_1^M {g(t)e^{ - 2t} \frac{{\sin t}}{t}dt} } \right| \ll \int_1^M {e^{ - 2t} dt} \ll 1$$ and $$\left| {\int_1^M {f(t)\frac{1}{{t^3 }}\tanh t\sin tdt} } \right| \ll \int_1^M {\frac{1}{{t^3 }}dt} \ll 1$$ uniformly in $$M$$. Thus, your improper integral converges if and only if the improper integral $$\int_1^{ + \infty } {\frac{{\sin t}}{t}dt}$$ converges.

Let \begin{align*} I_1 &= \int_1^\infty \frac{\sin t}{t} \,\mathrm{d} t, \\ I_2 &= \int_1^\infty \left(\frac{1}{t} - \arctan \frac{1}{t}\right)\sin t \,\mathrm{d} t, \\ I_3 &= \int_1^\infty \arctan \frac{1}{t} \cdot (1 - \tanh t) \cdot \sin t \,\mathrm{d} t. \end{align*} It is easy to prove that all $$I_1, I_2, I_3$$ exist (finite). See the remarks at the end.

Thus, $$I_1 - I_2 - I_3 = \int_1^\infty \arctan \frac{1}{t} \cdot \tanh t \cdot \sin t \, \mathrm{d} t$$ exists (finite).

We are done.

Remarks:

First, clearly $$I_1$$ exists (finite). Actually, $$I_1 = \int_0^\infty \frac{\sin t}{t}\, \mathrm{d} t - \int_0^1 \frac{\sin t}{t}\, \mathrm{d} t = \frac{\pi}{2} - \int_0^1 \frac{\sin t}{t}\, \mathrm{d} t$$.

Second, since $$\frac{1}{t} - \arctan \frac{1}{t} \ge 0$$ for all $$t \ge 1$$, we have $$\int_1^\infty \left|\left(\frac{1}{t} - \arctan \frac{1}{t}\right)\sin t\right| \,\mathrm{d} t \le \int_1^\infty \left(\frac{1}{t} - \arctan \frac{1}{t}\right) \,\mathrm{d} t = \frac{\pi}{4} + \frac{1}{2}\ln 2 - 1.$$ Thus, $$I_2$$ exists (finite).

Third, since $$1 - \tanh t \ge 0$$ for all $$t \ge 1$$ and $$0 \le \arctan \frac{1}{t} \le \frac{\pi}{4}$$ for all $$t \ge 1$$, we have $$\int_1^\infty \left|\arctan \frac{1}{t} \cdot (1 - \tanh t) \cdot \sin t \right|\,\mathrm{d} t \le \int_1^\infty \frac{\pi}{4} \cdot (1 - \tanh t) \,\mathrm{d} t = \frac{\pi}{4}[\ln(\mathrm{e} + \mathrm{e}^{-1}) - 1].$$ Thus, $$I_3$$ exists (finite).