What an Integral!!

The substitutions $u = e^t$ and $v = u^{-1}$ give $f(x) \; = \; \int_0^\infty e^{-\frac14x(u - u^{-1})^2}\,du \; = \; \int_0^\infty e^{-\frac14(v - v^{-1})^2}v^{-2}\,dv$ for $x > 0$ , and hence $f(x) \; = \; \tfrac12\int_0^\infty e^{-\frac14x(u - u^{-1})^2}\,\big(1 + u^{-2}\big)\,du \; = \; \tfrac12\int_{-\infty}^\infty e^{-\frac14xw^2}\,dw \; = \; \sqrt{\tfrac{\pi}{x}}$ for $x > 0$ , using the substitution $w = u - u^{-1}$ . Thus $f'(x) \; = \; -\frac{\sqrt{\pi}}{2x\sqrt{x}} \hspace{2cm} x > 0$ and so $f'(\pi) = -\tfrac{1}{2\pi}$ , making the answer $\boxed{\tfrac{1}{2\pi}}$ .