Prove that $\int_{0}^{\infty} \int_{0}^{1} \frac{\sin (\pi x)}{\pi} e^{-t} t^{-x} d t d x=e+\int_{0}^{1} \frac{e^{-t}}{\pi^{2}+\log ^{2}(t)} d t$

Question

Here's the story: I've been trying, without any success, for days now to derive the difference formula for the Fransén-Robinson constant. I was able to resolve one part, but then the equality up there is what is stopping me right now, from finishing off the calculations. I would highly appreciate some help. As of right now, I tried using Laplace transforms and Fubini to solve it, but the factor of $+e$ is always missing. I suppose it has to do with convergence issues.

Answer 1

I suppose that there is a typo in the double integral, $dtdx$ instead of $dxdt$ because if not the double integral will not be convergent. So, the presumed relationship might be : $$\int_{0}^{\infty} \int_{0}^{1} \frac{\sin (\pi x)}{\pi} e^{-t} t^{-x} d x d t \color{red}{\stackrel{?}{=}} e+\int_{0}^{1} \frac{e^{-t}}{\pi^{2}+\log ^{2}(t)} d t \tag 1$$

$\int_{0}^{1} \sin (\pi x) t^{-x} dx =\pi\frac{t+1}{t}\frac{1}{\ln^2(t)+\pi^2}$

$$\int_{0}^{\infty} \int_{0}^{1} \frac{\sin (\pi x)}{\pi} e^{-t} t^{-x} d x d t = \int_{0}^{\infty} \frac{t+1}{t}\frac{e^{-t}}{\pi^{2}+\log ^{2}(t)} d t$$

$$\int_{0}^{\infty} \int_{0}^{1} \frac{\sin (\pi x)}{\pi} e^{-t} t^{-x} d x d t = \int_{0}^{\infty} \frac{e^{-t}}{t(\pi^{2}+\log ^{2}(t))} d t + \int_{0}^{\infty} \frac{e^{-t}}{\pi^{2}+\log ^{2}(t)} d t$$

$$\int_{0}^{\infty} \int_{0}^{1} \frac{\sin (\pi x)}{\pi} e^{-t} t^{-x} d x d t \simeq 0.451747+0.089488 \simeq 0.541235$$ This is far from $$e+\int_{0}^{1} \frac{e^{-t}}{\pi^{2}+\log ^{2}(t)} d t \simeq 2.718282+0.053992\simeq2.772274 $$

Thus Eq.$(1)$ isn't correct. One cannot say why without checking the previous calculus which lead to this equation.

Moreover the original typing $\int_{0}^{\infty} \int_{0}^{1} \frac{\sin (\pi x)}{\pi} e^{-t} t^{-x} d t d x $ is obviously not correct because not convergent for $x\geq 1$ and $t\to 0$.