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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?

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Hint: $\ln(x+2)-\ln(x+1)$ is about $1/x$ (using mid-value theorem), for large $x$ –  Lior B-S Nov 28 '12 at 7:47
    
@LiorB-S: What is the mid-value theorem? I don't understand your hint. –  Samuel Reid Nov 28 '12 at 7:50
    
Sorry I made a literal translation. I meant the mean value theorem, en.wikipedia.org/wiki/Mean_value_theorem –  Lior B-S Nov 28 '12 at 7:57

4 Answers 4

up vote 2 down vote accepted

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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That is very helpful, thank you! –  Samuel Reid Nov 28 '12 at 7:58

By the mean value theorem $\ln\left(\frac{x+2}{x+1}\right)=\ln(x+2)-\ln(x+1) = \frac{1}{c(x)}$, where $x+1< c(x)<x+2$. Thus $\ln\left(\frac{x+2}{x+1}\right)\geq \frac{1}{x+2}$.

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.

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+1 Like this way. –  Babak S. Nov 28 '12 at 8:13

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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Another $+1^{{+1}^{+1^{10K}}}$ for my dear sleepy friend! –  amWhy Apr 7 '13 at 12:52

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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Can you explain the first equality? –  Lior B-S Nov 28 '12 at 15:45
    
@LiorB-S: Sure, the integral of log is a standard integral, $\int\log(t)dt = t\log t - t$. –  user26872 Nov 28 '12 at 18:17
    
ok thx. I think I get it, although, I haven't check the calculation... –  Lior B-S Nov 28 '12 at 19:20
    
@LiorB-S: You're welcome. I've added a few details. –  user26872 Nov 28 '12 at 19:48

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