I recently ran across the following integral:
$$ \int_{0}^{\infty}\frac{1}{\Gamma(x)}dx $$
Which I learned is equal to the Fransén-Robinson constant. On the linked wikipedia page for the Fransén-Robinson constant, it states that the difference between Fransén-Robinson constant and Euler's number can be expressed by this:
$$ F = e + \int_{0}^{\infty}\frac{e^{-x}}{\pi^2+\ln^2(x)}dx $$
Where on earth did the difference come from? How do we know this?
It is a consequence of the $\Gamma$ reflection formula: $$ \Gamma(z)\,\Gamma(1-z) = \frac{\pi}{\sin(\pi z)} \tag{1} $$ and the Cantarini's trick (aka the Laplace transform of the sine function): $$ \int_{0}^{+\infty} \sin(a t)\,e^{-bt} = \frac{a}{a^2+b^2}\tag{2} $$ from which:
$$ \frac{1}{\pi^2+\log^2(x)} = \int_{0}^{+\infty} \frac{\sin(\pi t)}{\pi} x^{-t}\,dt \qquad \left(\log(x)>0\right)\\\frac{1}{\pi^2+\log^2(x)} = \int_{0}^{+\infty} \frac{\sin(\pi t)}{\pi} x^{t}\,dt \qquad \left(\log(x)<0\right)\tag{3}$$ so $ \int_{0}^{+\infty}\frac{e^{-x}}{\pi^2+\log^2(x)}\,dx$ is related with: $$ \int_{0}^{+\infty}\frac{\sin(\pi t)}{\pi}\,\Gamma(1-t)\,dt = \int_{0}^{+\infty}\frac{dt}{\Gamma(t)}\tag{4}$$
Here is the process I found: $$\int_0^{\infty} \frac{1}{\Gamma(x)} dx = \int_0^{\infty} \frac{\Gamma(1-x)}{\pi} \sin{(\pi x)} dx = \int_0^{\infty} \frac{\sin{(\pi x)}}{\pi} \int_0^{\infty} e^{-t} t^{-x} dt dx$$ $$ = \int_0^\infty \frac{\sin{(\pi x)}}{\pi} \int_0^1 e^{-t} t^{-x} dt dx + \int_0^\infty \frac{\sin{(\pi x)}}{\pi} \int_1^\infty e^{-t} t^{-x} dt dx \equiv \mathcal{I}_1 + \mathcal{I}_2$$ We'll start evaluating the easiest one, $\mathcal{I}_2$: $$\mathcal{I}_2 = \int_0^\infty \frac{\sin{(\pi x)}}{\pi} \int_0^{\infty} e^{-t} t^{-x} dt dx = \int_1^\infty \int_0^{\infty} \frac{\sin{(\pi x)}}{\pi} e^{-t} e^{-x \log{t}} dx dt = \int_1^\infty \frac{e^{-t}}{{\pi}^2 + \log^2(t)} dt$$ Which constitutes a consistent part of the result. To obtain the rest, we shall be a bit more careful when exchanging integral signs. $$\mathcal{I}_1 = \int_0^\infty \frac{\sin{(\pi x)}}{\pi} \int_0^1 e^{-t} t^{-x} dt dx = \sum_{k \ge 0} \frac{(-1)^k}{k!} \int_0^\infty \frac{\sin{(\pi x)}}{\pi} \int_0^1 t^{k-x} dt dx $$ Thanks to the size of $t$, the sum converges and so do the integrals by themselves, we can go on interchanging and solving: $$= \sum_{k \ge 0} \frac{(-1)^k}{k!} \int_0^\infty \frac{\sin{(\pi x)}}{\pi (k-x+1)} dx = \frac{1}{\pi} \sum_{k \ge 0} \frac{(-1)^k}{k!} \int_{-(k+1)}^\infty \frac{(-1)^{k+1} \sin{(\pi x)}}{-x} dx$$ $$= \frac{1}{\pi} \sum_{k \ge 0} \frac{1}{k!} \int_{-(k+1)}^\infty \frac{\sin{(\pi x)}}{x} dx = \frac{1}{\pi} \sum_{k \ge 0} \frac{1}{k!} \int_{-\pi (k+1)}^\infty \frac{\sin{(x)}}{x} dx$$ Now break up the integral and compute the sum: $$= \frac{1}{\pi} \sum_{k \ge 0} \frac{1}{k!} \left[ \int_0^\infty \frac{\sin{(x)}}{x} dx + \int_{-\pi (k+1)}^0 \frac{\sin{(x)}}{x} dx \right] = \frac{e}{2} + \frac{1}{\pi} \sum_{k \ge 0} \frac{1}{k!} \int_0^{\pi (k+1)} \frac{\sin{(x)}}{x} dx $$ $$ = \frac{e}{2} + \frac{1}{\pi} \sum_{k \ge 0} \frac{1}{k!} \int_0^{\pi} \frac{\sin{((k+1)x)}}{x} dx = \frac{e}{2} + \frac{1}{\pi} \mathcal{Im} \left[ \int_0^{\pi} \frac{e^{ix}}{x} e^{e^{ix}} dx \right]$$ (the sum $\sum_{k \ge 0} \frac{\sin{(kx)}}{k!}$ can be easily evaluated by noticing $\sin{x} = \mathcal{Im} [e^{ix}]$ and using the definition of the exponential) $$ = \frac{e}{2} + \frac{1}{\pi} \mathcal{Im} \left[ \int_0^{\pi i} \frac{e^{y}}{y} e^{e^{y}} dy \right] = \frac{e}{2} - \frac{1}{\pi} \mathcal{Im} \left[ \int_{-1}^1 \frac{e^{x}}{\log{x}} dx \right] = \frac{e}{2} - \frac{1}{\pi} \mathcal{Im} \left[ \int_{-1}^1 \frac{e^{x}}{\ln{|x|} + i\arg{x}} dx \right]$$ We'll evaluate the argument of a pure negative number as $\pi i$, and take the limit as the argument of the pure real positive part goes to zero: $$ = \frac{e}{2} - \frac{1}{\pi} \int_{-1}^0 \mathcal{Im} \left[ \frac{e^{x}}{\ln{(-x)} + i\pi} \right] dx + \frac{1}{\pi} \int_0^1 \frac{e^{x}\arg{x}}{\ln^2{x} + i\arg^2{x}} dx$$ $$ = \frac{e}{2} + \frac{1}{\pi} \int_0^1 \frac{\pi e^{-x}}{\ln^2{x} + \pi^2} dx + \lim_{h\to 0} \frac{1}{\pi} \int_0^1 \frac{h e^{x}}{\ln^2{x} + h^2} dx$$ The limit will be solved by noting that the integrated function vanishes (for $h$ small enough) everywhere except a neighborhood of 1, hence the logarithm's and exponential's local behaviours predominate: $$ = \frac{e}{2} + \int_0^1 \frac{e^{-x}}{\ln^2{x} + \pi^2} dx + \frac{1}{\pi} \lim_{h\to 0} \int_{\mathcal{f}{(h)} \lt 1-h}^1 \frac{h e}{(1-x)^2 + h^2} dx = \frac{e}{2} + \int_0^1 \frac{e^{-x}}{\ln^2{x} + \pi^2} dx + \frac{1}{\pi} \frac{e \pi}{2}$$ $$= e + \int_0^1 \frac{e^{-x}}{\ln^2{x} + \pi^2} dx$$ Together with the first integral: $$\mathcal{F} = \mathcal{I}_1 + \mathcal{I}_2 = e + \int_0^\infty \frac{e^{-x}}{\ln^2{x} + \pi^2} dx$$
According to $(1)$ $$\frac{1}{\Gamma(z)} = \frac{\sin(\pi z)}{\pi}\Gamma(1-z)\tag{1}$$ And this naturally leads to $(4)$ $$ \int_{0}^{+\infty}\frac{\sin(\pi t)}{\pi}\,\Gamma(1-t)\,dt = \int_{0}^{+\infty}\frac{dt}{\Gamma(t)}\tag{4}$$ According to $(3)$, when $\ln(x)>0$ $$\frac{1}{\pi^2+\ln^2(x)} = \int_{0}^{\infty}\frac{\sin(\pi t)}{\pi}x^{-t}dt\tag{3}$$ I don't know if my following work is right or not. $$\int_{0}^{\infty}\frac{e^{-x}}{\pi^2+\ln^2(x)}dx = \int_{0}^{\infty}e^{-x}\int_{0}^{\infty}\frac{\sin(\pi t)}{\pi}x^{-t}dtdx\tag{a}$$ $$\int_{0}^{\infty}\frac{e^{-x}}{\pi^2+\ln^2(x)}dx = \int_{0}^{\infty}e^{-x}x^{-t}\int_{0}^{\infty}\frac{\sin(\pi t)}{\pi}dtdx\tag{b}$$ $$\int_{0}^{\infty}\frac{e^{-x}}{\pi^2+\ln^2(x)}dx = \int_{0}^{\infty}\frac{\sin(\pi t)}{\pi}\Gamma(1-t)dt\tag{c} = \int_{0}^{\infty}\frac{dt}{\Gamma(t)}$$ However, $$\int_{0}^{\infty}\frac{dt}{\Gamma(t)}= e + \int_{0}^{\infty}\frac{e^{-x}}{\pi^2+\ln^2(x)}dx $$ Where is the extra $e$?
