How to evaluate $\int_1^\infty \frac{1}{x}-\frac{1}{x+1}~dx$ It's a very simple question but it confuses me. How do I evaluate
$$
\int_1^\infty \frac{1}{x}-\frac{1}{x+1}~dx
$$
without splitting? And why can't I split it?
 A: You can't, because $x\mapsto \frac 1 x$ is not in $L_1$ and thus cannot be integrated over $[1,\infty)$ (as a matter of fact $\int_{[1,\infty)}\frac {dx} x = +\infty$).
However, $\frac 1 x-\frac 1 {x+1}$ can be rewritten (by reducing to the same denominator) into something which is in $L_1$ (and whose integral is easy to compute).
A: If I understand you correctly, you want
$$\begin{align}\int_1^{\infty} dx \left ( \frac{1}{x} - \frac{1}{x+1}\right ) &= \lim_{R \to \infty}\int_1^{R} dx \left ( \frac{1}{x} - \frac{1}{x+1}\right ) \\ &=\lim_{R \to \infty} [\log{R} - (\log{(R+1)} - \log{2})]\\ &= \log{2}  \end{align}$$
A: You can easily compute the integral of $\frac{1}{x}$ and $\frac{1}{x + 1}$ by substitution.  Then, you get logarithmic function.  But it is not defined at $\infty$.  Note that:
$$\int\limits_1^\infty \frac{1}{x} dx - \int\limits_1^\infty \frac{1}{x + 1} dx$$
$$= \lim_{t \rightarrow \infty} \int\limits_1^t \frac{1}{x} dx - \int\limits_1^t \frac{1}{x + 1} dx$$
For the second integrand, use substitution, letting $u = x + 1 \rightarrow du = dx$.  Then:
$$\lim_{t \rightarrow \infty} \int\limits_1^t \frac{1}{x} dx - \int\limits_1^t \frac{1}{u} u$$
$$= \lim_{t \rightarrow \infty} \ln{x} - \ln{u}|_{x = 1}^t$$
$$= \lim_{t \rightarrow \infty} \ln{x} - \ln{x + 1}|_{x = 1}^t$$
$$= \lim_{t \rightarrow \infty} \ln{t} - \ln{1} - (\ln(t + 1) - \ln(1 + 1))$$
So if we substitute $t$ with $\infty$, then we obtain the diverging integral.
