# Prove that the limit converges to $\gamma$

Let $$H(x)=\int_0^1\frac{t^x-1}{t-1}dt$$be the harmonic series and let $$s(x)=\int^\infty_0e^{-t}\ln(t+x)dt$$How do I prove that their difference converges? It seems to me that they approach $$\gamma$$ (Euler-Mascheroni Constant) according to this graph. But how would I go about proving this? I think we have to utilize that fact that $$-\gamma=\int_0^1H(x)dx$$and the fact that $$s(0)=-\gamma$$, but I don't know what else I could do.

This question was inspired by the fact that: $$\int_0^1e^{-t}\ln tdt=-\gamma\text{ and}\int_0^1e^{-t}\ln(t+1)dt=\delta$$Where $$\delta$$ is the Euler-Gompertz constant. I decided to generalize this to a function $$s(x)$$. When I graphed it I right away thought of relating it to the Harmonic series and the natural logarithm. Since it seems that $$s(x)\sim\ln x$$, the problem is equivalent to proving that: $$\lim_{x\rightarrow\infty}\int_0^\infty e^{-t}\ln(t+x)-\frac{e^{-tx}-e^{-t}}{t}dt=0$$ The random looking second term comes from the fact that $$\int_0^\infty\frac{e^{-tx}-e^{-t}}{t}dt=\ln x$$

• If you use that $H(x)=\ln(x)+\gamma+O(x^{-1})$, all you need to show is that $s(x)-\ln(x)=e^x\int_x^\infty\frac{e^{-t}}{t}dt$ goes to $0$. Jan 2, 2023 at 23:19

$$\newcommand{\d}{\,\mathrm{d}}$$Well, you know (typo in your OP): $$\int_0^\infty\ln(t)e^{-t}\d t=-\gamma$$And: $$s(x)=\int_0^\infty e^{-t}\ln(1+x/t)\d t+\int_0^\infty e^{-t}\ln(t)\d t=\sigma(x)-\gamma$$We just want to show the difference: $$H(x)-\sigma(x)$$Tends to zero as $$x\to\infty$$, in order to show: $$H(x)-s(x)\to\gamma$$We have:
$$H(x)-\sigma(x)=\int_0^\infty\left[H(x)-\ln(x)-\ln\left(\frac{1}{x}+\frac{1}{t}\right)\right]e^{-t}\d t$$
Pointwise, for some $$t>0$$ fixed, the above integrand is convergent to: $$(\gamma+\ln(t))e^{-t}$$For every $$t$$. Inspecting the proof of convergence, we can get some bounds and apply the dominated convergence theorem to say: $$\lim_{x\to\infty}H(x)-\sigma(x)=\int_0^\infty(\gamma+\ln(t))e^{-t}\d t=0$$
As desired. $$\blacksquare$$
Using the above link, we can say that: \begin{align}\left|H(x)-\ln(x)-\ln\left(\frac{1}{x}+\frac{1}{t}\right)\right|&<\sum_{j=1}^\infty\frac{1}{j(j+1)}+\left|\ln\left(\frac{1}{x}+\frac{1}{t}\right)\right|\\&=1+\left|\ln\left(\frac{t}{1+\frac{t}{x}}\right)\right|\\&<1+|\ln(t)|\end{align}And the dominated convergence theorem now clearly applies after multiplication with $$e^{-t}$$.