Compute $\lim_{x\to 0^+}\int_{2x}^{3x}\frac{\sin(t)}{\sinh^2(t)}dt$. How can I compute $$\lim_{x\to 0^+}\int_{2x}^{3x}\frac{\sin(t)}{\sinh^2(t)}dt \ \ ?$$
Suppose $x<\pi$. I tried using DCT since $$\left|\frac{\sin(t)}{\sinh^2(t)}\boldsymbol 1_{[2x,3x}(t)\right|\leq \frac{\sin(t)}{\sin^2(t)},$$
but the function on the RHS is not integrable... It should have a trick.
 A: You have$$\lim_{t\to0^+}\frac{\sin(t)}{\sinh^2(t)}-\frac1t=\lim_{t\to0^+}\frac{t\sin(t)-\sinh^2(t)}{t\sinh^2(t)}=0$$and therefore\begin{align}\lim_{x\to0^+}\int_{2x}^{3x}\frac{\sin(t)}{\sinh^2(t)}\,\mathrm dt&=\lim_{x\to0^+}\left(\int_{2x}^{3x}\frac{\sin(t)}{\sinh^2(t)}-\frac1t\,\mathrm dt\right)+\lim_{x\to0^+}\int_{2x}^{3x}\frac{\mathrm dt}t\\&=\lim_{x\to0^+}\bigl(\log(3x)-\log(2x)\bigr)\\&=\log\left(\frac32\right).\end{align}
A: The Taylor development of the smooth even function
$$\frac{t\sin t}{\sinh^2t}$$
is of the form
$$1-\frac{t^2}2+o(t^2)$$
so that for small $t$ there is certainly a finite $a$ such that
$$1-at^2\le\frac{t\sin t}{\sinh^2t}\le1.$$
(In fact $a=\dfrac12$ works.)
This implies
$$\frac1t-at\le\frac{\sin t}{\sinh^2t}\le\frac1t,$$
and integrating between $2x$ and $3x$ ($<t$),
$$\log\frac32-ax\le I\le \log\frac32.$$
A: Put $ f(t)=\dfrac{t\sin(t)}{\sinh^2(t)} $. Then
$\lim_{t\to 0^{+}}f(t) = 1$.
From the Second mean value theorem for definite integrals we get
\begin{equation*}
\int_{2x}^{3x}\dfrac{\sin(t)}{\sinh^2(t)}\, \mathrm{d}t = \int_{2x}^{3x}f(t)\dfrac{1}{t}\, \mathrm{d}t = f(\xi)\int_{2x}^{3x}\dfrac{1}{t}\, \mathrm{d}t = f(\xi)\log\left(\dfrac{3}{2}\right)
\end{equation*}
where $ 2x<\xi <3x$. Consequently
\begin{equation*}
\lim_{t\to 0^{+}}\int_{2x}^{3x}\dfrac{\sin(t)}{\sinh^2(t)}\, \mathrm{d}t = \log\left(\dfrac{3}{2}\right).
\end{equation*}
A: You could also use series expansion of the integrand. This would give
$$\frac{\sin(t)}{\sinh^2(t)}=\frac{t-\frac{t^3}{6}+\frac{t^5}{120}+O\left(t^7\right) }{\left(t+\frac{t^3}{6}+\frac{t^5}{120}+O\left(t^7\right) \right)^2 }=\frac{t-\frac{t^3}{6}+\frac{t^5}{120}+O\left(t^7\right) } {t^2+\frac{t^4}{3}+\frac{2 t^6}{45}+O\left(t^8\right) }=\frac{1}{t}-\frac{t}{2}+\frac{47 t^3}{360}+O\left(t^5\right)$$ Now, integrate, use the bounds and so on.
