Can these identities for the Euler-Mascheroni constant be proven?

Question

I stumbled upon these 4 limit/integral identities involving Euler's constant aka gamma (~0.5772). They appear to be valid based on inspection but I have no idea how to prove them. In addition, I have no idea how these identities connect to the commonly known identities for gamma, or even how any of them follow from connect to each other. Any insight would be much appreciated!

$$ \gamma = \lim_{x \to \infty } -x+ \frac{1}{\int_{0}^{1}\frac{t^{x-t}}{\Gamma(t)}dt} $$ $$ \gamma = \lim_{x \to \infty } -x+2+ \frac{1}{\int_{1}^{k}\frac{\Gamma(t)}{t^{x-t}}dt} $$ $$ -\gamma = \lim_{x \to \infty } -x-2+ \frac{1}{\int_{0}^{1}\frac{\Gamma(t)}{t^{-x-t}}dt} $$ $$ -\gamma = \lim_{x \to \infty } -x+ \frac{1}{\int_{1}^{k}\frac{t^{-x-t}}{\Gamma(t)}dt}. $$

$$k\in \mathbb{R}, k>1$$

Answer 1

$\newcommand\ga\gamma\newcommand\Ga\Gamma$Concerning the first question, we have to show that $$g(x):=-x+1/I(x)\to\ga \tag{1}\label{1}$$ (as $x\to\infty$), where $$I(x):=I_1(h,x)+I_2(h,x),$$ $$I_1(h,x):=\int_0^{1-h} dt\,\frac{t^{x-1}}{t^{t-1}\Ga(t)},\quad I_2(h,x):=\int_{1-h}^1 dt\,\frac{t^{x-1}}{t^{t-1}\Ga(t)},$$ $0<h<1$. Since $t^{t-1}\Ga(t)\ge c$ for some $c>0$ and all $t\in(0,1)$, we have $$I_1(h,x)=O\big(\int_0^{1-h} dt\,t^{x-1}\Big)=O((1-h)^x).$$

To estimate $I_2(h,x)$, note that $\frac1{t^{t-1}\Ga(t)}=1-(\ga+o(1))(1-t)$ as $t\uparrow1$. So, if $h\downarrow0$ so that $(1-h)^x=o(x^{-2})$ (say, if $h=1/\sqrt x$ for $x>1$), then $$I_2(h,x)=\int_{1-h}^1 dt\,t^{x-1}(1-(\ga+o(1))(1-t)) =\frac1x\,\Big(1-\frac{\ga+o(1)}x\Big)$$ and hence $$I(x)=\frac1x\,\Big(1-\frac{\ga+o(1)}x\Big)$$ and $$\frac1{I(x)}=x\,\Big(1+\frac{\ga+o(1)}x\Big)=x+\ga+o(1),$$ so that \eqref{1} follows.

The other formulas should follow similarly. (There should be only one question in one post.)

Answer 2

For a sufficiently regular function (such as $\Gamma(t+1)$ or its inverse),

$$\int_0^1 t^xf(t)\,dt=\dfrac{f(1)}{x+1}-\dfrac{f'(1)}{(x+1)(x+2)}+\cdots\;,$$ so your formulas follow from $\Gamma'(1)=-\gamma$

Answer 3

Let's look at the first identity, $$\gamma = \lim_{x \to \infty }\left( -x+ \frac{1}{\int_{0}^{1}\frac{t^{x-t}}{\Gamma(t)}dt}\right).$$ The integrand $t^{x-t}/\Gamma(t)$ is sharply peaked at $t=1$ for large $x$. I expand $$\frac{t^{-t}}{\Gamma(t)}=1+(t-1)(\gamma-1)+{\cal O}(t-1)^2,$$ and evaluate $$ \lim_{x \to \infty }\left( -x+ \frac{1}{\int_{0}^{1}\frac{t^{x-t}}{\Gamma(t)}dt}\right)=\lim_{x\to\infty}\left(-x+\frac{1}{\int_0^1 t^x[1+(t-1)(\gamma-1)]\,dt}\right)$$ $$\qquad=\lim_{x\to\infty}\frac{\gamma x+2}{x-\gamma +3}=\gamma.$$