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I am making a mistake and I don't quite see where. Consider the following integral $$ I\equiv\int_2^{\infty}\frac{e^{-x}-1}{x}\frac{\log\frac{1}{x}}{\log^2\frac{1}{x}+\frac{\pi^2}{4}}dx $$ I have plotted the integrand and the function $\frac{1}{x^{1.1}}$. My mathematica tells me that up to $x=10000$ the integrand above is smaller than $\frac{1}{x^{1.1}}$. It is well-known that $$ \int_2^{\infty}\frac{1}{x^{1.1}}dx $$ is convergent. Therefore, I expect my integral to be convergent.

Nonetheless, when I proceed with integration by parts I encounter infinities. This is how. Noticing $$ \int\frac{\log\frac{1}{x}}{x(\log^2\frac{1}{x}+\frac{\pi^2}{4})}dx=-\frac{1}{2}\log(\pi^2+4\log^2\frac{1}{x}) $$ I can write $$ I=(e^{-x}-1)\big(-\frac{1}{2}\log(\pi^2+4\log^2\frac{1}{x})\big)\bigg|_2^{\infty}-\frac{1}{2}\int_2^{\infty}e^{-x}\log(\pi^2+4\log^2\frac{1}{x}) $$ and we encounter an infinity in the first term above. Where did I go wrong? EDIT: picture of the plot in Mathematica. The blue line is the integrand and the orange one $1/x^{1.1}$enter image description here

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Hint. Your integral is divergent because, as $x \to \infty$, $$ \frac{e^{-x}-1}{x}\frac{\log\frac{1}{x}}{\log^2\frac{1}{x}+\frac{\pi^2}{4}} \sim\frac{-1}{x}\frac{\log\frac{1}{x}}{\log^2\frac{1}{x}} \sim \frac{1}{x\log x} $$ and the latter integrand gives a divergent integral over $[2,\infty)$. One may recall that, as $M \to \infty$, $$ \int_2^M \frac{1}{x\log x}\:dx =\left[\frac{}{}\log\left(\log x \right) \right]_2^M =\log\left(\log M\right) -\log\left(\log 2 \right) \to \infty. $$

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  • $\begingroup$ I thought about that, but why is it the integrand then smaller than $1/x^{1.1}$ always? $\endgroup$ Apr 4 '17 at 15:44
  • $\begingroup$ @AnarchistBirdsWorshipFungus Do you have any picture of it? $\endgroup$ Apr 4 '17 at 15:48
  • $\begingroup$ Plot added in the question $\endgroup$ Apr 4 '17 at 15:53
  • $\begingroup$ @AnarchistBirdsWorshipFungus Thanks. How do you know which one is the blue curve? $\endgroup$ Apr 4 '17 at 15:55
  • $\begingroup$ I have checked it $\endgroup$ Apr 4 '17 at 16:08
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When working with convergence questions, very often when logs are involved, graphing up to $x=100000$ is not good enough. Convergence/divergence only cares about what happens as $x\to\infty$.

You are essentially comparing $\log x$ and $x^{0.1}$. You know that asymptotically any positive power of $x$ is eventually larger than $\log x$. But consider that $\log 100000 \approx 11.5$, while $100000^{0.1} \approx 3.2$. This doesn't mean that what you know is wrong, it only means that $x = 100000$ is absolutely tiny compared to the scale that is relevant when comparing these two functions. Try $x=10^{100}$ and we see that $\log x \approx 230$ while $x^{0.1} = 10^{10}$ is obviously so much larger.

The moral here is if you need to determine behaviour at infinity, graphing up to $x=100000$ (or even $x=10^{100}$) will never be completely conclusive. You're only looking at only a microscopic sliver of the whole picture.

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