Hot Integral - 18

Since $\cosh(x+iy) \,=\, \cosh x \cos y + \sinh x \sin y$ , we see that $f(t) \; =\; \mathrm{Im}\left(\int_0^\pi \cosh\big(\tfrac{1+i}{\sqrt{2}}t\sin x\big)\,dx\right)$ and hence $\begin{array}{rcl} (\mathcal{L}f)(s) & = & \displaystyle \int_0^\infty \mathrm{Im}\left(\int_0^\pi \cosh \big(\tfrac{1+i}{\sqrt{2}}t\sin x\big)\,dx\right)\,e^{-st}\,dt \\ & = & \displaystyle\dfrac12\mathrm{Im}\left(\int_0^\pi \int_0^\infty \Big(e^{\frac{1+i}{\sqrt{2}}t\sin x} + e^{-\frac{1+i}{\sqrt{2}}t\sin x}\Big) e^{-st}\,dt\,dx\right) \\ & = & \displaystyle \dfrac12\mathrm{Im}\left(\int_0^\pi\left(\frac{1}{s-\frac{1+i}{\sqrt{2}}\sin x} + \frac{1}{s + \frac{1+i}{\sqrt{2}}\sin x}\right)\,dx\right) \\ & = & \displaystyle \mathrm{Im}\int_0^\pi \dfrac{s}{s^2 - i\sin^2x}\,dx \; = \; s\int_0^\pi \dfrac{\sin^2x}{s^4 + \sin^4x}\,dx \; = \; s\int_0^\pi \dfrac{\mathrm{cosec}^2\,x}{s^4\mathrm{cosec}^4\,x + 1}\,dx \\ & = & \displaystyle s\int_{\mathbb{R}} \frac{1}{s^4(t^2+1)^2 + 1}\,dt \; = \; \dfrac{1}{s^3}\int_{\mathbb{R}} \frac{1}{(t^2 + 1 + is^{-2})(t^2 + 1 - is^{-2})}\,dt \\ & = & \displaystyle\dfrac{1}{s^3}\int_{\mathbb{R}} \frac{1}{\big(t^2 + \sec \theta e^{i\theta}\big)\big(t^2 + \sec\theta e^{-i\theta}\big)}\,dt \end{array}$ where $\theta = \tan^{-1}s^{-2}$ and we are using the substitution $t = \cot x$ .

By standard contour integration, we need to evaluate the residues of this integrand at the poles with positive imaginary part. These are $i\sqrt{\sec\theta}e^{\pm\frac12i\theta}$ , and so $\begin{array}{rcl} (\mathcal{L}f)(s) & = & \displaystyle \dfrac{2\pi i}{s^3}\left(\mathrm{Res}_{z=i\sqrt{\sec\theta}e^{\frac12i\theta}} + \mathrm{Res}_{z=i\sqrt{\sec\theta}e^{-\frac12i\theta}}\right) \frac{1}{\big(z^2 + \sec \theta e^{i\theta}\big)\big(z^2 + \sec\theta e^{-i\theta}\big)} \\ & = & \displaystyle \dfrac{2\pi i}{s^3}\left\{ -\frac{1}{\sec\theta(e^{i\theta}-e^{-i\theta})}\frac{1}{2i\sqrt{\sec\theta}e^{\frac12i\theta}} + \frac{1}{\sec\theta(e^{i\theta}-e^{-i\theta})}\frac{1}{2i\sqrt{\sec\theta}e^{-\frac12i\theta}}\right\} \\ & = & \dfrac{\pi}{2s^3} \dfrac{(\cos\theta)^{\frac32}}{\cos\frac12\theta} \end{array}$ So much for the integration. Now to get the result in the right shape to answer the question. Since $\tan\theta = s^{-2}$ , we have $\sec^2\theta =\dfrac{s^4+1}{s^4}$ , and so $\cos\theta = \dfrac{s^2}{\sqrt{s^4+1}}$ . Thus $(\mathcal{L}f)(s) \; = \; \frac{\pi}{2(s^4+1)^{\frac34}} \sec\tfrac12\theta \;.$ Finally we have $\sin\theta = \dfrac{1}{\sqrt{s^4 + 1}}$ , and hence $(\mathcal{L}f)(s) \; = \; \frac{\pi}{2(s^4+1)^{\frac14}}\sin\theta \sec\tfrac12\theta \; = \; \frac{\pi}{(s^4+1)^{\frac14}}\sin\tfrac12\theta \; = \; \frac{\pi}{(s^4+1)^{\frac14}}\sin\left[\tfrac12\tan^{-1}\big(s^{-2}\big)\right] \;.$ Thus the answer is $1+4+1+1+4+1 + 2 + 1 +2 + 1 = 18$ .