I was trying to evaluate this integral: $$\int_{0}^{+\infty}e^{-x}H_{0}^{(1)}(x)dx$$ (here $H_{0}^{(1)}$ denotes the Hankel function of the first kind.
Wolfram Alpha returns me $$\frac{\pi-2i\text{arcsinh}(1)}{\sqrt{2}\pi}.$$I have no idea on how it gets this closed form (in fact consulting Abramovitz allowed me to verify the real part of the value, but no clue about the rest).
Thanks for your help.
We rewrite $H_{0}^{(1)}(x)$ in terms of Bessel functions to get $$\int_{0}^{\infty}e^{-x}H_{0}^{(1)}(x) \ dx = \int_{0}^{\infty}e^{-x}J_{0}(x) \ dx + i\int_{0}^{\infty}e^{-x}Y_{0}(x) \ dx$$ We can view the first integral as the Laplace Transform of $J_{0}(x)$ evaluated at $s = 1$. The proof is here. We have that $$\int_{0}^{\infty}e^{-x}J_{0}(x) \ dx = \mathcal{L}\{J_{0}(x)\}[1] = \frac{1}{\sqrt{2}}$$ We now look at the integral $$\int_{0}^{\infty}e^{-x}Y_{0}(x) \ dx$$ Using 10.9.9, we rewrite as $$-\frac{2}{\pi}\int_{0}^{\infty}e^{-x}\int_{0}^{\infty}\cos(x\cosh(t)) \ dt \ dx$$ $$-\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}e^{-x}\cos(x\cosh(t)) \ dt \ dx$$ $$-\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}e^{-x}\cos(x\cosh(t)) \ dx \ dt$$ $$-\frac{2}{\pi}\int_{0}^{\infty}\frac{1}{\cosh^{2}{t}+1} \ dt$$ $$-\frac{2}{\pi}\frac{\ln{(3 + 2^{3/2})} - \ln{(3 - 2^{3/2})}}{2^{5/2}}$$ We bring it all together to get $$\int_{0}^{\infty}e^{-x}H_{0}^{(1)}(x) \ dx = \frac{1}{\sqrt{2}} - i\frac{2}{\pi}\frac{\ln{(3 + 2^{3/2})} - \ln{(3 - 2^{3/2})}}{2^{5/2}}$$ which is equivalent to $$\frac{\pi - 2i\text{arcsinh}(1)}{\sqrt{2} \pi}$$