4
$\begingroup$

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 $$\int_0^\infty\frac{e^{-tx}-e^{-t}}{t}dt=\ln x$$

$\endgroup$
1
  • $\begingroup$ 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 $0$. $\endgroup$ Jan 2, 2023 at 23:19

1 Answer 1

3
$\begingroup$

$\newcommand{\d}{\,\mathrm{d}}$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 $e^{-t}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .