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I came across this limit while integrating Bessel functions: $$\lim_{b\to\infty}b\int_0^1\cos(b x) \cosh^{-1}(\frac{1}{x})dx$$

This integral does not have any standard value I know of for fixed $b$. Graphing it numerically, it appears to converge to $\pi/2$ as $b\to\infty$ (albeit slowly).

Graph of the value of the integral as b increases.

Does anyone have any ideas for evaluating such an integral/limit?

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By integration by parts $$ \int_{0}^{1}b\cos(bx)\cosh^{-1}\left(\frac{1}{x}\right)\,dx = \left.\sin(bx)\log\left(\frac{1+\sqrt{1-x^2}}{x}\right)\right|_{0^+}^{1}+\int_{0}^{1}\frac{\sin(bx)}{x\sqrt{1-x^2}}\,dx$$ that equals $$ \int_{0}^{b}\frac{\sin(x)}{x\sqrt{1-\frac{x^2}{b^2}}}\,dx $$ and by the dominated convergence theorem, as $b\to +\infty$ the last integral approaches $$ \int_{0}^{+\infty}\frac{\sin(x)}{x}\,dx = \color{red}{\frac{\pi}{2}} $$ as expected.

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  • $\begingroup$ How does one apply the dominated convergence theorem here? It requires the magnitude of the integrand to be integrable, which is not the case for $\frac{|\sin(x)|}{x}$ on $[0,\infty)$. For instance, $\lim_\limits{s\to 0}\int_0^\infty e^{-s t} \cos(t)dt=1$ whereas $\int_0^\infty \cos(t)dt$ does not converge. $\endgroup$ – Grant B. Mar 30 '17 at 5:21
  • $\begingroup$ Excuse me, $\lim_\limits{s\to 0} \int_0^\infty e^{-st}\cos(t)dt = 0$. $\endgroup$ – Grant B. Mar 30 '17 at 5:28
  • $\begingroup$ @GrantB.: obviously $\frac{\sin x}{x}$ is not a Lebesgue integrable function over $\mathbb{R}^+$, but it is an improperly Riemann-integrable function.You may switch to Laplace transform or just apply the DCT to $\frac{\sin x}{x}\mathbb{1}_{(0,b)}(x)$. $\endgroup$ – Jack D'Aurizio Mar 30 '17 at 9:06

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