I have a reduced a problem I have been working on to showing that $\int_{0}^{\infty} \ln \left(\frac{x+2}{x+1} \right)dx$ diverges, but I'm not sure how to show this. Would anyone be able to help me out?
Divergence of a simple improper integral $\int_{0}^{\infty} \ln \left(\frac{x+2}{x+1} \right)dx$
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0Sorry I made a literal translation. I meant the mean value theorem, http://en.wikipedia.org/wiki/Mean_value_theorem – 2012-11-28
4 Answers
A pedestrian approach: \begin{eqnarray*} \int_{0}^{n} \ln \left(\frac{x+2}{x+1} \right)dx &=& \int_{0}^{n} \ln \left(x+2\right)dx -\int_{0}^{n} \ln \left(x+1\right)dx \\ &=& \ldots \\ &=& \log\frac{(n+2)^{n+2}}{4(n+1)^{n+1}} \\ &=& \log n + O(1) \hspace{5ex} (\textrm{expand in large }n) \end{eqnarray*} Above we use the standard integral $\int\log(t)dt = t\log t - t.$
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0@LiorB-S: You're welcome. I've added a few details. – 2012-11-28
By the mean value theorem $\ln\left(\frac{x+2}{x+1}\right)=\ln(x+2)-\ln(x+1) = \frac{1}{c(x)}$, where $x+1< c(x)
Since the integral $\int_{1}^\infty \frac{1}{x+2} dx$ diverges, we get that $\int_{1}^\infty \ln\left(\frac{x+2}{x+1}\right)dx$ diverges, and hence $\int_{0}^\infty \ln\left(\frac{x+2}{x+1}\right)dx$ diverges.
First note $\ln\left(\dfrac{x+2}{x+1} \right)$ is a decreasing function. Hence, $\int_{n-1}^{n} \ln\left(\dfrac{x+2}{x+1} \right) dx > \ln \left(\dfrac{n+2}{n+1} \right)$ Hence, \begin{align} \int_0^{N} \ln\left(\dfrac{x+2}{x+1} \right) dx & = \sum_{n=1}^{N} \int_{n-1}^{n} \ln\left(\dfrac{x+2}{x+1} \right) dx\\ & > \sum_{n=1}^{N} \ln \left(\dfrac{n+2}{n+1} \right)\\ & = \sum_{n=1}^{N} \left(\ln(n+2) - \ln(n+1) \right)\\ & = \ln(N+2) - \ln(2) \end{align} Now you should be able to finish it off by letting $N \to \infty$.
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0That is very helpful, thank you! – 2012-11-28
Another approach may arise as following:
Let $g(x)=\frac{1}{x+1}$. Then it is non-negative and $\lim_{x\to +\infty}\frac{\ln\frac{x+2}{x+1}}{\frac{1}{x+1}}=1$ which is not zero or $\infty$; so the Quotient test says, $\int_0^{\infty}\ln\frac{x+2}{x+1} dx $ and $\int_0^{\infty}\frac{1}{x+1} dx $ are the same in being divergence or in being convergence. The second integral is clearely divergent one.
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0Another $+1^{{+1}^{+1^{10K}}}$ for my dear sleepy friend! – 2013-04-07