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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.

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  • $\begingroup$ The integrand is asymptotic to $1/t$. $\endgroup$ – Yves Daoust Jul 18 '18 at 10:20
  • $\begingroup$ For the fun of it, try numerical integration with $x=\frac 14$; you should get $\approx 0.335004$ while, using the expansion, you should get $\approx 0.335627$. $\endgroup$ – Claude Leibovici Jul 18 '18 at 10:36
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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}

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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 $3x$ ($<t$),

$$\log\frac32-ax\le I\le \log\frac32.$$

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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*}

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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.

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