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When trying to derive Mertens' theorem using Perron's formula, I encountered the following integral:

$$ \int_0^1{1-e^{-t}\over t}\mathrm dt-\int_1^\infty{e^{-t}\over t}\mathrm dt $$

which, according to Titchmarsh's The Theory of Riemann Zeta Function, evaluates to $\gamma$ (a.k.a. Euler-Mascheroni constant).

To begin with, I converted this integral into the following limit:

$$ \lim_{N\to\infty}\left[\int_0^1{1-(1-t/N)^N\over t}\mathrm dt-\int_1^N{(1-t/N)^N\over t}\mathrm dt\right] $$

so that I can eventually simplify the expression into

$$ \lim_{N\to\infty}\left[\sum_{m=1}^N\frac1m-\log N\right] $$

which is identical to the definition of $\gamma$.

I wonder what tools can be used to justify the limit step.

TravorLZH
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1 Answers1

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METHODOLOGY $1$:

To show that

$$\lim_{N\to\infty}\int_1^N \frac{\left(1-\frac tN\right)^N}{t}\,dt=\int_1^\infty \frac{e^{-t}}{t}\,dt$$

we will show that

$$\lim_{N\to\infty}\int_1^N \frac{e^{-t}}t \left(e^t\left(1-\frac tN\right)^N-1\right)\,dt=0\tag 1$$

Proceeding, we use the inequalities

$$\left(1-\frac tN\right)^N\le e^{-t}\tag 2$$

and

$$\begin{align} e^t\left(1-\frac tN\right)^N &\ge \left(1+\frac tN\right)^N\left(1-\frac tN\right)^N\\\\ &=\left(1-\frac{t^2}{N^2}\right)^N\\\\ &\ge 1-\frac{t^2}{N}\tag3 \end{align}$$

Using $(2)$ and $(3)$, we have the estimates for the integral in $(1)$

$$-\frac1N \int_1^N te^{-t}\,dt \le \int_1^N \frac{e^{-t}}t \left(e^t\left(1-\frac tN\right)^N-1\right)\,dt\le 0\tag4$$

whence application of the squeeze theorem to $(4)$ yields $(1)$.

To show that

$$\lim_{N\to\infty}\int_0^1 \frac{1-\left(1-\frac tN\right)^N}{t}\,dt=\int_0^1 \frac{1-e^{-t}}{t}\,dt$$

we exploit the uniform convergence of $\left(1-\frac tN\right)^N$ to $e^{-t}$ for $t\in [0,1]$. Then, we may interchange the limit and integral to obtain the result.

Putting it together, we have

$$\lim_{N\to \infty}\left(\int_1^N \frac{\left(1-\frac tN\right)^N}{t}\,dt-\int_0^1 \frac{1-\left(1-\frac tN\right)^N}{t}\,dt\right)=\int_0^1 \frac{1-e^{-t}}{t}\,dt-\int_1^\infty \frac{e^{-t}}{t}\,dt$$

as was to be shown.

METHODOLOGY $2$:

There is another way to proceed. We can write the integral of interest as

$$\begin{align} I&=\int_0^1 \frac{1-e^{-t}}{t}\,dt-\int_1^\infty \frac{e^{-t}}{t}\,dt\\\\ &=\lim_{\varepsilon\to 0^+}\int_\varepsilon^1 \frac{1-e^{-t}}{t}\,dt-\lim_{L\to \infty}\int_1^L \frac{e^{-t}}{t}\,dt\tag1 \end{align}$$

Integrating by parts both of the integrals on the right-hand side of $(1)$ we find

$$\begin{align} I&=\lim_{\varepsilon\to 0^+}\left(-(1-e^{-\varepsilon})\log(\varepsilon)-\int_\varepsilon^1 e^{-t}\log(t)\,dt\right)\\\\ &-\lim_{L\to \infty}\left(e^{-L}\log(L)-\int_1^L e^{-t}\log(t)\,dt\right)\\\\ &=-\int_0^\infty e^{-t}\log(t)\,dt\\\\ \end{align}$$

Finally, using the development in the NOTE section of THIS ANSWER, we see that $-\int_0^\infty e^{-t}\log(t)\,dt=\gamma$ as was to be shown!

metamorphy
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Mark Viola
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    This presumes that one knows the final integral representation. But the OP's definition is in terms of a limit, so converting from one integral representation to another doesn't exactly seem like progress. – Semiclassical Mar 12 '21 at 16:43
  • @Semiclassical The resulting integral representation herein is a standard representation. In fact, in the "NOTE" section of [THIS ANSWER}(https://math.stackexchange.com/questions/2100567/real-analysis-methodologies-to-show-gamma-2-int-0-infty-frac-cosx2-cos/2101763#2101763), I showed that it is equal to the limit definition in the OP. – Mark Viola Mar 12 '21 at 17:03
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    @Semiclassical I've edited to address the OP's question and kept, as a second approach, my initial answer. ;-) – Mark Viola Mar 12 '21 at 18:14
  • Well, the second method is what I was trying to avoid since I am writing an article about this for my classmates and do not want to get too much into Gamma function. Still, +1 for your elegant first method :D – TravorLZH Mar 13 '21 at 06:14