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)
Best Answer
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.