Integral $\int_{0}^{+\infty}\frac{1-e^{-x}-\log(1+x)}{x^2}dx$.

Question

$$\int_{0}^{+\infty}\frac{1-e^{-x}-\log(1+x)}{x^2}dx$$

Though of Using Maclaurin for log but I don't think it will get me anywhere and nor does the 1/x substitution work. Any hint?

Answer 1

As with Jack D'Aurizio's solution, we begin by integrating by parts with $u=1-e^{-x}-\log(1+x)$ and $v=\frac1x$ to obtain

$$\begin{align} \int_0^\infty \frac{1-e^{-x}-\log(1+x)}{x^2}\,dx&=\int_0^\infty \frac{e^{-x}-\frac1{1+x}}{x}\,dx\\\\ &=\lim_{\epsilon\to0^+}\left(\int_\epsilon^\infty \frac{e^{-x}}{x}\,dx-\int_\epsilon^\infty \frac{1}{x(1+x)}\,dx\right)\tag1 \end{align}$$

We integrate by parts the first integral on the right-hand side of $(1)$ to reveal

$$\begin{align} \int_0^\infty \frac{1-e^{-x}-\log(1+x)}{x^2}\,dx&=\lim_{\epsilon\to0^+}\left(e^{-\epsilon}\log(\epsilon)+\int_\epsilon^\infty e^{-x}\log(x)\,dx-\log(\epsilon)-\log(1+\epsilon)\right)\\\\ &=\int_0^\infty e^{-x}\log(x)\,dx\tag2\\\\ &=-\gamma \end{align}$$

And we are done!

---

NOTE: I showed in the Note at the end of THIS ANSWER, the equivalence of the integral representation of

$$-\gamma=\int_0^\infty e^{-x}\log(x)\,dx$$

in $(2)$ and the limit representation

$$-\gamma=\lim_{N\to \infty}\left(-\log(N)+\sum_{n=1}^N\frac{1}{n}\right)$$

Answer 2

Hint : $\frac {\frac x{1+x} - ln(1+x)}{x^2} = \frac d{dx} \{ \frac {ln(1+x)}x \}$

Then write it as $\int_{0}^ \infty \frac {1-e^{-x} - {\frac x{1+x}}}{x^2} + $ $\int_{0}^ \infty \ d \left (\frac {ln(1+x)}x \right ) $

Answer 3

The issue in using the Maclaurin series for $\log(1+x)$ is that its radius of convergence is one, while the integration range extends past $1$. On the other hand, by integration by parts, the given integral equals $$ \int_{0}^{+\infty}\left(e^{-x}-\frac{1}{x+1}\right)\frac{dx}{x} $$ which is a well-known integral representation for the opposite of the Euler-Mascheroni constant, $\color{blue}{-\gamma}$.

Here it is a proof through the usual definition of $\gamma$. A useful lemma is $\int_{0}^{+\infty}\frac{e^{-ax}-e^{-bx}}{x}\,dx = \log\frac{b}{a}$ for any $a,b>0$, which is Frullani's integral. We have

$$ \gamma=\lim_{n\to +\infty} H_n-\log n = \sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right]$$ hence the RHS can be represented as $$ \int_{0}^{+\infty}\sum_{n\geq 1} e^{-nx}-\frac{e^{-nx}-e^{-(n+1)x}}{x}\,dx =\int_{0}^{+\infty}\left(\frac{1}{x e^x}-\frac{1}{e^x-1}\right)\,dx$$ and it is enough to show that $$ \int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{x(x+1)}\right)\,dx=0, $$ which follows from $\int\frac{dx}{x(x+1)}=C+\log(x)-\log(1+x)$ and $\int\frac{dx}{e^x-1}=C+\log(e^x-1)-x.$