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)

  • 4
    $\begingroup$ I don't know what sen is. $\endgroup$ Jul 25, 2021 at 11:14
  • $\begingroup$ Apparently, sen stands for “seno” which is Spanish for “sine”. A simple search reveals this. $\endgroup$ Jul 25, 2021 at 11:43
  • $\begingroup$ @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.) $\endgroup$ Jul 25, 2021 at 11:51
  • $\begingroup$ If you read this board a lot, you will see "sen" from time to time. $\endgroup$
    – GEdgar
    Jul 25, 2021 at 12:01
  • 1
    $\begingroup$ 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 $\endgroup$
    – irbag
    Jul 25, 2021 at 12:24

2 Answers 2


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.


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).


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