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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?

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1  
A good place to start: notice the relationship between the bounds of the first and second integrals. In particular, $\ln 2 = -\ln {1 \over 2}$ and $\ln 3 = -\ln {1 \over 3}$. This suggests the substitution $u = \ln t$. –  Théophile May 30 '12 at 1:35
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And remember $${\tanh ^{ - 1}}x = \frac{1}{2}\log \left( {\frac{{1 + x}}{{1 - x}}} \right)$$ –  Pedro Tamaroff May 30 '12 at 1:37
    
Ohh okay, so you have to do it by substitution? Thanks, didnt realise that! Will try working on it.. –  Steven May 30 '12 at 1:45
    
@JoeL., (I just learned myself that) $\operatorname{artanh}$ is a standard name for $\tanh^{-1}$ (see, e.g., wiki), so we should probably stick to Steven's original spelling. –  Antonio Vargas May 30 '12 at 16:05
    
@AntonioVargas You are correct - thanks for the catch. I had just assumed that he had made a typo and there should be a "c." –  Joe May 30 '12 at 16:42

2 Answers 2

up vote 3 down vote accepted

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.

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Thanks Brian, most helpful! –  Steven May 30 '12 at 2:10

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!

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