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The above identity is the difference formula for the Fransén-Robinson Constant. Proving this statement gave me severe headaches those last days, since everytime I try to calculate the RHS I either miss the $+e$ term or nothing converges. In the picture you can find one of my attempts. As it seems as of now, the integral in the last line on the left does not converge. I'm completely at a loss. I would appreciate some help alot!!

My attempt 3:

Flammable Maths
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  • this link might help you: https://math.stackexchange.com/questions/1802176/how-was-the-difference-of-the-frans%C3%A9n-robinson-constant-and-eulers-number-found – Ricardo770 Jul 07 '19 at 11:13
  • They go through the same process basically^^ The person asking the question is encoiuntering the same problem with the missing e XD – Flammable Maths Jul 07 '19 at 11:17
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    Related. Set $(\alpha, \beta, t) = (-1, 1, 1)$. – Maxim Jul 07 '19 at 18:26
  • The integral formula $\Gamma(z) = \int_0^\infty t^{z-1}e^{-t},dt$ works only for $\operatorname{Re}z>0,$ since that's where the integral converges absolutely. Full $\Gamma$ is then defined as analytic continuation. In particular, $\Gamma(1-x) = \int_0^\infty t^{-x}e^{-t},dt$ works for $x<1$ and the integral doesn't converge otherwise. – Ennar Jul 08 '19 at 16:29

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I found the following proof in the book Ramanujan by G. H. Hardy, which Hardy attributes to Ramanujan.

Define $$f(x,y) = \int_{-y}^\infty\frac{x^t}{\Gamma(1+t)}\,dt + \int_0^\infty t^{y-1}\frac{e^{-xt}}{\pi^2+\log^2t}(\cos(\pi y)-\frac{\sin(\pi y)}{\pi}\log t)\,dt,\ x\geq 0,\, y\geq 0$$ and first differentiate it wrt to $y$ to get \begin{align} \frac{\partial f}{\partial y}(x,y) &= \frac{x^{-y}}{\Gamma(1-y)}+\int_0^\infty\frac{e^{-xt}}{\pi^2+\log^2t}\frac{\partial}{\partial y}\left[t^{y-1}(\cos(\pi y)-\frac{\sin(\pi y)}{\pi}\log t)\right]\,dt \\ &= \frac{x^{-y}}{\Gamma(1-y)}-\int_0^\infty\frac{e^{-xt}}{\pi^2+\log^2t}t^{y-1}(\pi^2+\log^2t)\frac{\sin(\pi y)}{\pi}\,dt \\ &= \frac{x^{-y}}{\Gamma(1-y)}-\frac{\sin(\pi y)}{\pi}\int_0^\infty e^{-xt}t^{y-1}\,dt \\ &= \frac{x^{-y}}{\Gamma(1-y)}-\frac{\sin(\pi y)}{\pi} x^{-y}\Gamma(y) = 0 \end{align} where the last equality is due to Euler's reflection formula.

Thus, $f$ doesn't depend on $y$.

Now, differentiate $f$ wrt to $x$ to get \begin{align} \frac{\partial f}{\partial x}(x,y) &= \int_{-y}^\infty\frac{x^{t-1}}{\Gamma(t)}\,dt - \int_0^\infty t^{y}\frac{e^{-xt}}{\pi^2+\log^2t}(\cos(\pi y)-\frac{\sin(\pi y)}{\pi}\log t)\,dt \\ &= \int_{-(y+1)}^\infty\frac{x^{t}}{\Gamma(t+1)}\,dt - \int_0^\infty t^{(y+1)-1}\frac{e^{-xt}}{\pi^2+\log^2t}(-\cos(\pi (y+1))+\frac{\sin(\pi (y+1))}{\pi}\log t)\,dt \\ &= f(x,y+1) = f(x,y). \end{align} It follows that $f(x,y) = Ce^x$, and by plugging in $x = y = 0$, we get $C = 1$. Thus, letting $y = 0$ we get integral formula $$\int_{0}^\infty\frac{x^t}{\Gamma(1+t)}\,dt = e^x - \int_0^\infty \frac{e^{-xt}}{t(\pi^2+\log^2t)}\, dt$$ which was known to Ramanujan and used (and generalized) by Hardy in some of his papers.

All you have to do now is differentiate the last integral formula wrt $x$ and let $x = 1$ to obtain the desired result.

Ennar
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    Awesome! But.. why in particular that function? Would there be any intuition behind taking it? I know it's Ramanujan, but still.. – Zacky Jul 11 '19 at 13:12
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    @Zacky, well, try to prove the last formula I wrote. Obviously, an idea could be to differentiate integrals and show the result is the same function, which is enough to conclude that it is exponential function. That's what I tried at first. However, when you just differentiate that, the result is not obvious to be equal to the begging expression so I was stuck. Ramanujan's brilliance is to introduce this new variable such that the whole expression is independent of it, but can be used to show that the expression is derivative of itself. And like usual, who the heck knows how he thought of it. – Ennar Jul 11 '19 at 13:20
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    @Zacky, that last formula is not hard to prove using inverse Laplace transform and residue theorem. But anyway, it seems that Ramanujan was the first to think of it, and again, how is not clear. What is clear is that the LHS should be approximately $e^x$ just by replacing the integral by series. – Ennar Jul 11 '19 at 13:28
  • @Ennar Could you please share your thoughts on the following? If we set $x = 0$ in the formula $$\int_{0}^\infty\frac{x^t}{\Gamma(1+t)},dt = e^x - \int_0^\infty \frac{e^{-xt}}{t(\pi^2+\log^2t)}, dt,$$ the result is $0=1-1$, which is fine. But if we differentiate w.r.t. to $x$, and if differentiation under the integral is possible, one would get $$\int_{0}^\infty\frac{x^{t-1}}{\Gamma(t)},dt = e^x + \int_0^\infty \frac{e^{-xt}}{(\pi^2+\log^2t)}, dt,$$ and setting $x=0$ gives $0=1+\infty$. Where does this problem come from? – Holden Jan 09 '20 at 16:18
  • @Holden, $t-1$ can obtain negative values so you end up with singularity at $x = 0$. – Ennar Jan 09 '20 at 17:31
  • @Ennar but does then the derivation of the formula remain valid? – Holden Jan 09 '20 at 17:57
  • @Ennar, maybe one could start the derivation by taking the function $f(x,y)$ for $x \color\red{>0}$, $y \geq 0$. Then obtain that $f(x, y) = C e^x$ for $x>0$ and $y \geq 0$, which means that $$\int_{0}^\infty\frac{x^t}{\Gamma(1+t)},dt - \int_0^\infty \frac{e^{-xt}}{t(\pi^2+\log^2t)}, dt= Ce^x$$ and then... – Holden Jan 09 '20 at 18:26
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    @Holden, I didn't really check the details, what I wrote is almost verbatim from Hardy. Looks like for $x>0$ there is no issue whatsoever in the derivation, so you end up with $f(x,y) = Ce^x$ for $x>0$. Now, if you let $y=0$ at the beginning, there is no issue for letting $x = 0$, and we only do that to calculate $C$. Now just insert some topological/analytic continuation argument to justify extension of the formula. – Ennar Jan 09 '20 at 18:26
  • @Ennar, $\int_{0}^\infty\frac{x^t}{\Gamma(1+t)},dt$ and $\int_0^\infty \frac{e^{-xt}}{t(\pi^2+\log^2t)}$ are respectively equal to $0$ and $1$ at $x=0$, and are probably right-continuous. Then it seems that making $x \downarrow0$ in the formula $$\int_{0}^\infty\frac{x^t}{\Gamma(1+t)},dt + \int_0^\infty \frac{e^{-xt}} {t(\pi^2+\log^2t)}, dt= Ce^x \quad x>0$$ should yield $C=1$. Also, there should be a "+" instead of a "-" in my previous comment. – Holden Jan 09 '20 at 18:56
  • I am having a problem. If I let $E(x)=\int_0^{\infty}x^t/\Gamma(1+t)dt,$ then $E'(x)=\int_0^\infty tx^{t-1}/\Gamma(1+t)dt=\int_0^\infty x^{t-1}/\Gamma(t)dt=\int_{-1}^0x^t/\Gamma(1+t)dt+E(x).$ Letting $t\mapsto -t$ and using the $\Gamma$ reflection formula, $$E'(x)-E(x)=\frac{1}{\pi}\int_0^1x^{-t}\Gamma(t)\sin(\pi t)dt$$ – clathratus Feb 10 '23 at 01:32
  • Then using the integral definition for $\Gamma(t)$ and switching the order of integration, $$E'(x)-E(x)=\frac{1}{\pi}\int_0^\infty\frac{e^{-s}}{s}\int_0^1\left(\frac{s}{x}\right)^t\sin(\pi t)dt ds.$$ Then the inner integral is $$\int_0^1\left(\frac{s}{x}\right)^t\sin(\pi t)dt=\frac{\pi}{\pi^2+\log^2(s/x)},$$ so that $$E'(x)-E(x)=\int_0^\infty\frac{e^{-s}ds}{s(\pi^2+\log^2(s/x))}=\int_0^\infty\frac{e^{-xt}dt}{t(\pi^2+\log^2t)}$$ – clathratus Feb 10 '23 at 01:38
  • then using your method we clearly have $$E(x)=e^x-\int_0^\infty\frac{e^{-xt}dt}{t(\pi^2+\log^2t)}.$$ Combining this with my formula for $E'-E$, we have $$E(x)=e^x-(E'(x)-E(x)),$$ i.e. $E'(x)=e^x$ so that $E(x)=e^x$. Assuming your derivation is correct, where did mine go wrong? – clathratus Feb 10 '23 at 01:43
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    @clathratus, it should be $\int_0^1\left(\frac{s}{x}\right)^t\sin(\pi t)dt=\frac{\pi(s+x)}{x(\pi^2+\log^2(s/x))}$, no? – Ennar Feb 10 '23 at 03:47
  • @Ennar oh my god what a silly mistake! Thanks for catching it I was going crazy – clathratus Feb 10 '23 at 04:28