Intermediate step in deriving integral representation of Euler–Mascheroni constant: $\int_0^1\frac{1-e^{-t}-e^{-1/t}}{t}dt$

Question

I'm following a complex analysis course and am making an exercise in which I have to derive an integral representation for the Euler–Mascheroni constant.

I have the following definition of the Euler–Mascheroni constant: $$\gamma = \lim_{n\rightarrow\infty}\left(1 + 1/2 + +\ldots+ 1/n - \log n\right)$$

I have shown (with induction): $$1 + 1/2 + \ldots+ 1/n = \int_0^1\frac{1-(1-t)^n}{t}dt$$

I now have to show: $$\gamma = \lim_\limits{n\to\infty}\left(\int_0^1 (1-(1-t/n)^n)\frac{dt}{t} - \int_1^n(1-t/n)^n\frac{dt}{t}\right)$$

By using $e^t = \lim\limits_{n\to\infty}(1+t/n)^n$ and substituting $t$ for $1/t$ in the second integral, it then follows: $$\gamma = \int_0^1\frac{1-e^{-t}-e^{-1/t}}{t}dt$$

None of my ideas have been promising for the second step. Any suggestions?

Answer 1

@metamorphy pointed to this answer in the comments. It is an answer of a question about the same integral representation of $\gamma$ but with different intermediate steps. I adapted it to the answer below.

We have: $$1 + 1/2 + \ldots+ 1/n = \int_0^1\frac{1-(1-t)^n}{t}dt$$ By making the change of variable $t = x/n$, we get: $$\int_0^1\frac{1-(1-t)^n}{t}dt = \int_0^n\frac{1-(1-x/n)^n}{x}dx$$ So:

\begin{align} \gamma & = \lim_{n\rightarrow\infty}\left(1 + 1/2 + +\ldots+ 1/n - \log n\right)\\ & = \lim_{n\rightarrow\infty}\left(\int_0^n\frac{1-(1-x/n)^n}{x}dx - \int_1^n\frac{1}{x}dx\right)\\ & = \lim_{n\rightarrow\infty}\left(\int_0^1\frac{1-(1-x/n)^n}{x}dx + \int_1^n\frac{1-(1-x/n)^n}{x}dx - \int_1^n\frac{1}{x}dx\right)\\ & = \lim_{n\rightarrow\infty}\left(\int_0^11-(1-x/n)^n\frac{dx}{x} - \int_1^n(1-x/n)^n\frac{dx}{x} \right)\\ \end{align}

Answer 2

Note that the integral in question can be split into two separately covergent pieces that after a trivial substitution read:

$$I=\int_0^1\frac{1-e^{-t}-e^{-1/t}}{t}dt=\int_0^1\frac{1-e^{-t}}{t}dt- \int_0^1\frac{e^{-1/t}}{t}dt=\int_0^1\frac{1-e^{-t}}{t}dt- \int_1^\infty\frac{e^{-u}}{u}du$$

We notice that we can combine the 2nd term of the first integral on the LHS with the second integral, but alas, the 2nd term of the first integral does not converge on it's own, so we cannot apply the linearity of the integral to split it. To salvage the applicability of the rule we replace the problematic lower bound by a finite parameter:

$$I=\lim_{\epsilon\to 0}\left(\int_\epsilon^1\frac{1-e^{-t}}{t}dt- \int_1^\infty\frac{e^{-u}}{u}du\right)=\lim_{\epsilon\to 0}\left(\int_\epsilon^1\frac{1}{t}dt- \int_\epsilon^\infty\frac{e^{-u}}{u}du\right)$$

Performing an integration by parts on the second term we get

$$I=\lim_{\epsilon\to 0}\left(-(1-e^{-\epsilon})\log \epsilon- \int_\epsilon^\infty e^{-u}\log u du\right)=-\int_0^\infty e^{-t}\log t dt=\gamma$$

The last equation is a well known representation of the Euler-Mascheroni constant. A proof that connects the definition with this integral representation is given here.