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