Use comparison test to determine convergence

Question

$$\int_{1}^{\infty}\frac{\ln x}{\sinh x}dx$$ I tried several functions and failed to get integrable convergent bigger function.

Thanks for help.

Answer 1

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.

Answer 2

Cauchy-Schwarz is enough:

$$\begin{eqnarray*} \int_{1}^{M}\frac{\log x}{\sinh x}\,dx &\leq&\sqrt{\int_{1}^{M}\frac{\log^2 x}{x^2}\,dx \cdot \int_{1}^{M}\frac{x^2}{\sinh^2 x}\,dx}\\&\leq&\sqrt{\int_{1}^{+\infty}\frac{\log^2 x}{x^2}\,dx\cdot\int_{0}^{+\infty}\frac{x^2}{\sinh^2 x}\,dx}\\&=&\sqrt{2\cdot\frac{\pi^2}{6}}=\frac{\pi}{\sqrt{3}}.\end{eqnarray*} $$

Answer 3

$$0~<~\int_1^\infty\frac{\ln x}{\sinh x}~dx~<~\int_1^\infty\frac{x}{\sinh x}~dx~<~\int_0^\infty\frac{x}{\sinh x}~dx~=~\bigg(\frac\pi2\bigg)^2$$

Answer 4

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 $$ \begin{align} \int_1^\infty\frac{\log(x)}{\sinh(x)}\,\mathrm{d}x &\le\frac{2e}{e-1}\int_1^\infty\frac{\log(x)}{e^x}\,\mathrm{d}x\\ &=\frac{2e}{e-1}\int_1^\infty\frac{e^{-x}}{x}\,\mathrm{d}x\\ &\le\frac{2e}{e-1}\int_1^\infty e^{-x}\,\mathrm{d}x\\ &=\frac2{e-1} \end{align} $$