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I have a lot of doubts about improper integrals with logarithm. In those integrals in which there are other functions apart from the logarithm usually is ok. If there is just the logarithm it is a bit harder for me. For example considering

$$\int_{1}^{\infty }(Log[x+1]-Log[x])dx$$ $$\forall x\in [1,\infty)\, \, \, \, \, \, \, (Log[x+1]-Log[x])=Log\left [\frac{x+1}{x} \right ] $$ $$Log\left [\frac{x+1}{e^{x}} \right ]\leq Log\left [\frac{x+1}{x} \right ]\, \, \, \, \forall x\in [1,\infty)$$ $$\int_{1}^{\infty }(Log\left [\frac{x+1}{e^{x}} \right ])dx=\int_{1}^{\infty }(-x+Log\left [{x+1} \right ])dx$$ $$ x \mapsto \infty \, \, \, \, \, \, \, -x+Log[x+1]\sim -x $$

$$\int_{1}^{\infty }-x \,dx$$ The integral is divergent. So for comparison theorem for improper integrals $$\int_{1}^{\infty }(Log[x+1]-Log[x])dx$$ should be divergent. Is it correct? Is the correct choice in those cases to use the exponential in the logarithm to create an inequality that when later we will evaluate the second improper integral leaves the logarithm . Do you have any techniques useful in those cases?

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With $\ds{\Lambda > 1}$:

\begin{align} &\color{#f00}{\int_{1}^{\Lambda}\bracks{\ln\pars{x + 1} - \ln\pars{x}}\,\dd x} = \int_{2}^{\Lambda + 1}\ln\pars{x}\,\dd x - \int_{1}^{\Lambda}\ln\pars{x}\,\dd x \\[5mm] = &\ \bracks{\pars{\Lambda + 1}\ln\pars{\Lambda + 1} - \pars{\Lambda + 1} - 2\ln\pars{2} + 2} - \bracks{\Lambda\ln\pars{\Lambda} - \Lambda} \\[5mm] & = 1 - 2\ln\pars{2} +\ \underbrace{\Lambda\ln\pars{1 + {1 \over \Lambda}}} _{\ds{\to\ 1\ \mbox{as}\ \Lambda\ \to\ \infty}}\ + \underbrace{\ln\pars{\Lambda + 1}}_{\ds{\to\ \color{#f00}{\infty}\ \mbox{as}\ \Lambda\ \to\ \infty}}\quad \color{#f00}{\to \infty}\ \mbox{as}\ \color{#f00}{\Lambda \to \infty} \end{align}

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Why not simply compute the antidericative $$\int \log(y)\,dy=y \log (y)-y$$ So $$\int \left( \log(1+x)-\log(x)\right)\,dx=-x \log (x)+(x+1) \log (x+1)$$ So $$f(t)=\int_1^t \left( \log(1+x)-\log(x)\right)\,dx=(t+1) \log (t+1)-t \log (t)-2\log(2)$$ $$f(t)=(t+1)\left(\log(t)+\log(1+\frac 1t)\right)-t \log (t)-2\log(2)$$ $$f(t)=(t+1)\log(1+\frac 1t)+\log(t)-2\log(2)$$ Now using Taylor $$\log(1+\frac 1t)=\frac{1}{t}-\frac{1}{2 t^2}+O\left(\frac{1}{t^3}\right)$$ $$(t+1)\log(1+\frac 1t)=1+\frac{1}{2 t}+O\left(\frac{1}{t^2}\right)$$ $$f(t)=1-2\log(2)+\log(t)+\frac{1}{2 t}+O\left(\frac{1}{t^2}\right)$$

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Your argument doesn't do the job since $\int_1^{b} -x\,dx \to -\infty$ as $b \to \infty$ and hence you have shown that $-\infty \leq \int_1^{\infty} (\log(x+1)-\log(x))\,dx$, which we could all agree on before your argument.

Perhaps use a different bounding argument. By the Mean Value Theorem, for each $x \in (1, \infty)$ there is some $c_x \in [x, x+1]$ such that $\log(x+1) - \log(x) = \log'(c_x)(x+1-x) = \frac{1}{c_x} \in \left[\frac{1}{x+1}, \frac{1}{x}\right]$ and hence, $\frac{1}{x+1} \leq \log(x+1)-\log(x) \leq \frac{1}{x}$.

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  • $\begingroup$ $\frac 1x>\frac1{x+1}$. $\endgroup$ – Mark Viola Aug 27 '16 at 18:20
  • $\begingroup$ @Dr.MV Totally correct! Thanks for the reminder. $\endgroup$ – Tom Aug 27 '16 at 18:21
  • $\begingroup$ You're welcome Tom. My pleasure. -Mark $\endgroup$ – Mark Viola Aug 27 '16 at 18:24
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In THIS ANSWER, I showed using only Bernoulli's Inequality and the limit definition of the exponential function that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le\log(x)\le x-1 \tag 1$$

Therefore, using $(1)$ we find

$$\frac{1}{x+1} \le \log\left(1+\frac1x\right)\le \frac{1}{x}$$

Since, the integral $\int_1^\infty \frac{1}{x+1}\,dx$ diverges, the integral of interest does also.

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