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)
 A: 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.
A: 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).
