4
$\begingroup$

How would I show (or explain) that

$\int_\frac{1}{3}^\frac{1}{2}\frac{\operatorname{artanh} t}{t}dt,$ $\int_{\ln 2}^{\ln 3}\frac{u}{2\sinh u}du,$ and $-\int_\frac{1}{3}^\frac{1}{2}\frac{\ln v}{1-v^2}dv$ are all equivalent, without evaluating any of the integrals?

  • 0
    @AntonioVargas You are correct - thanks for the catch. I had just assumed that he had made a typo and there should be a "c."2012-05-30

2 Answers 2

3

As Théophile suggested in the comments, the limits of integration provide a natural starting point: $\ln\frac12=-\ln 2$, $\ln\frac13=-\ln 3$, and $-\int_\frac{1}{3}^\frac{1}{2}\frac{\ln v}{1-v^2}dv=\int_\frac{1}{2}^\frac{1}{3}\frac{\ln v}{1-v^2}dv\;,$ so one should consider the possibility that $u=-\ln v$. If so, $du=-\frac1v dv$, and $\sinh u=\frac12\left(e^u-e^{-u}\right)=\frac12\left(e^{-\ln v}-e^{\ln v}\right)=\frac12\left(\frac1v-v\right)=\frac{1-v^2}{2v}\;,$ so

$\int_{\ln 2}^{\ln 3}\frac{u}{2\sinh u}du=\int_{\frac12}^{\frac13}\frac{-\ln v}{\frac{1-v^2}v}\left(-\frac1v\right)dv=\int_{\frac13}^{\frac12}\frac{\ln v}{1-v^2}dv=-\int_{\frac12}^{\frac13}\frac{\ln v}{1-v^2}dv\;.$

The other equalities will also succumb to reasonable substitutions.

  • 0
    Thanks Brian, most helpful!2012-05-30
2

I'd like to give a suggestion for the equality

$\displaystyle \int_{1/3}^{1/2} \frac{\text{arctanh } t}{t}\text{ d}t=-\int_{1/3}^{1/2} \frac{\log v}{1-v^2}\text{ d}v$

The idea is to rewrite $\displaystyle \text{arctanh } t=\int_0^t \frac{1}{1-s^2}\text{ d}s$, so that we have

$\displaystyle \int_{1/3}^{1/2} \frac{\text{arctanh } t}{t}\text{ d}t=\int_{1/3}^{1/2} \int_0^t \frac{1}{1-s^2} \frac{1}{t} \text{ d}s \text{ d}t$

Now we can change the order of integration (note that the region isn't "simple"):

$\displaystyle \int_{1/3}^{1/2} \int_0^t \frac{1}{1-s^2} \frac{1}{t} \text{ d}s \text{ d}t=\int_{0}^{1/3} \int_{1/3}^{1/2} \frac{1}{1-s^2} \frac{1}{t} \text{ d}t \text{ d}s+\int_{1/3}^{1/2} \int_{s}^{1/2} \frac{1}{1-s^2} \frac{1}{t} \text{ d}t \text{ d}s$

It is very easy to explicitly evaluate several of the terms now:

$\begin{align*} &=\displaystyle \frac{1}{2} \log(3/2) \log(2)+\int_{1/3}^{1/2} \frac{\log(1/2)-\log s}{1-s^2}\text{ d}s \\ &=\frac{1}{2} \log(3/2) \log(2)+\int_{1/3}^{1/2} \frac{\log(1/2)}{1-s^2}\text{ d}s - \int_{1/3}^{1/2} \frac{\log s}{1-s^2}\text{ d}s \\ &=\frac{1}{2} \log(3/2) \log(2)-\frac{1}{2} \log(3/2) \log(2) - \int_{1/3}^{1/2} \frac{\log s}{1-s^2}\text{ d}s \\ &=- \int_{1/3}^{1/2} \frac{\log s}{1-s^2}\text{ d}s \end{align*}$

This is precisely the desired term!