Use comparison test to determine convergence $$\int_{1}^{\infty}\frac{\ln x}{\sinh x}dx$$
I tried several functions and failed to get integrable convergent bigger function.
Thanks for help.
 A: Since the integrand is continuous on $[1,+\infty)$, a potential problem is at $x \to +\infty$.
As suggested by André Nicolas, using
$$
\sinh x > x^3/6,\qquad x>0,
$$
gives
$$
\begin{align}
\int_{1}^{\infty}\frac{\ln x}{\sinh x}dx&<6\int_{1}^{\infty}\frac{\ln x}{x^3}dx\\\\
&=\left.-\frac{3}{x^2} \ln x\right|_1^{\infty}+3\int_{1}^{\infty}\frac1{x^3}dx\\\\
&=3 \left.\left(-\frac{1}{2 x^2}\right)\right|_1^{\infty}\\\\
&=\frac32.
\end{align}
$$ Your integral is then convergent.
A: For $x\ge0$, $e^{-x}\le1$, therefore,
$$
\begin{align}
\sinh(x)
&=\frac{e^x-e^{-x}}2\\
&\ge\frac{e^x-1}2
\end{align}
$$
and for $x\ge1$, $e^x/e\ge 1$. Therefore,
$$
\begin{align}
\frac{e^x-1}{2}
&\ge\frac{e^x-e^x/e}2\\
&=e^x\frac{e-1}{2e}
\end{align}
$$
Thus, the previous estimates and integration by parts yields
$$
\begin{align}
\int_1^\infty\frac{\log(x)}{\sinh(x)}\,\mathrm{d}x
&\le\frac{2e}{e-1}\int_1^\infty\frac{\log(x)}{e^x}\,\mathrm{d}x\\
&=\frac{2e}{e-1}\int_1^\infty\frac{e^{-x}}{x}\,\mathrm{d}x\\
&\le\frac{2e}{e-1}\int_1^\infty e^{-x}\,\mathrm{d}x\\
&=\frac2{e-1}
\end{align}
$$
A: Cauchy-Schwarz is enough:
$$\begin{eqnarray*} \int_{1}^{M}\frac{\log x}{\sinh x}\,dx &\leq&\sqrt{\int_{1}^{M}\frac{\log^2 x}{x^2}\,dx \cdot \int_{1}^{M}\frac{x^2}{\sinh^2 x}\,dx}\\&\leq&\sqrt{\int_{1}^{+\infty}\frac{\log^2 x}{x^2}\,dx\cdot\int_{0}^{+\infty}\frac{x^2}{\sinh^2 x}\,dx}\\&=&\sqrt{2\cdot\frac{\pi^2}{6}}=\frac{\pi}{\sqrt{3}}.\end{eqnarray*} $$
A: 
$$0~<~\int_1^\infty\frac{\ln x}{\sinh x}~dx~<~\int_1^\infty\frac{x}{\sinh x}~dx~<~\int_0^\infty\frac{x}{\sinh x}~dx~=~\bigg(\frac\pi2\bigg)^2$$

