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Check the convergence of the following integral and, if it converges, compute its value: $$\int_0^\infty \frac{\ln\left(\frac{1+x^{(\sqrt3+2)}}{1-x^{(\sqrt3-2)}}\right)}{(1+x^2)\ln x} \, dx$$

Efforts:

$$\int_0^\infty \frac{1}{1+x^2} \ln_x\left(\frac{1+x^{\sqrt3+2}}{1-x^{\sqrt3-2}}\right) \,dx$$

I couldn't find any common feature, $1-x^{(\sqrt3-2)}$ with $1+x^{(\sqrt3+2)}$

And I can't understand how can be "$x^\sqrt3$" like things.

$$\to \int_0^\infty \frac{1}{1+x^2} \cdot \ln_x\left(\frac{1+x^{\sqrt3+2}}{1-x^{\sqrt3-2}}\right) \, dx= \int_0^\infty \dfrac{1}{1+x^2} \cdot \ln_x \left( \dfrac{1+x^\sqrt3\cdot x^2}{\dfrac{x^2-x^\sqrt3}{x^2}}\right) \, dx$$

$$=\int_0^\infty \frac{1}{1+x^2}\cdot\ln_x\left(\frac{x^2+x^\sqrt3\cdot x^4}{x^2-x^\sqrt3}\right) \, dx$$

I also tried to use the substitution: $u=x^{x^2+1}$

$$du=\left[x^{(x^2)}+x^{(x^2+2)}+\ln\left(x^{(2x+2x^3)}\right)\right] \, dx$$

Here too I was not able to prove anything.

Unfortunately I can't find the solution, please give a hint or an answer.

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    $\begingroup$ I'm not sure how to do the integral, but notice that the manipulation $1 - x^{\sqrt3 - 2} = \dfrac{1 - x^{\sqrt3}}{x^2}$ is not correct. $\endgroup$ – LSpice Jun 5 '16 at 16:35
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    $\begingroup$ Mathematica says that the integral diverges. $\endgroup$ – LSpice Jun 5 '16 at 16:56
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    $\begingroup$ Also, I think that the change in the title is a shame; now the title gives no information about the integral of interest. (A rough glance at questions tagged 'integration' suggests that it is common to include at least a description of the integral in question.) $\endgroup$ – LSpice Jun 5 '16 at 17:07
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    $\begingroup$ The argument of $\ln$ in the integrand is negative on the interval $(0,1)$.@L Spice $\endgroup$ – John Wayland Bales Jun 5 '16 at 18:27
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    $\begingroup$ desmos.com/calculator/wj6vwmpews $\endgroup$ – John Wayland Bales Jun 5 '16 at 19:20
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Let $$I(\alpha)=\int_{0}^{+\infty}\frac{\log(1+x^\alpha)-\log(2)}{(1+x^2)\log(x)}\,dx.\tag{1}$$ Then: $$ I'(\alpha) = \int_{0}^{+\infty}\frac{x^{\alpha}}{(1+x^2)(1+x^{\alpha})}\,dx =\frac{\pi}{2}-\int_{0}^{+\infty}\frac{dx}{(1+x^2)(1+x^{\alpha})}\tag{2}$$ where the last integral does not really depend on $\alpha$: it is enough to split the integration range into $(0,1)\cup(1,+\infty)$ and apply the substitution $x\mapsto\frac{1}{x}$ on the second interval to see a wonderful cancellation happening. So, simply, $I'(\alpha)=\frac{\pi}{4}$ and $\color{red}{I(\alpha)=\frac{\pi\alpha}{4}}$. On the other hand, by setting: $$ J(\beta) = \int_{0}^{+\infty}\frac{\log(2)-\log(1-x^\beta)}{(1+x^2)\log x}\,dx \tag{3}$$ we have a non-integrable singularity at $x=1$, hence the original integral cannot be convergent.

Maybe a typo somewhere?

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    $\begingroup$ @LSpice: $I(\alpha)$ is a convergent integral, there is only a singularity at the origin, but it is a integrable singularity. $\endgroup$ – Jack D'Aurizio Jun 5 '16 at 17:10
  • $\begingroup$ I apologise; I read in a hurry and thought that $I(\alpha)$ was the original integral. I will delete my comment. $\endgroup$ – LSpice Jun 5 '16 at 17:11
  • $\begingroup$ Thanks a lot of.to everyone.. $\endgroup$ – user2312512851 Jun 5 '16 at 17:18

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