Expert Answer to "Show:∫1312arctanh(t)tdt=∫ln⁡2ln⁡3u2sinhudu"

Cosmo Truong

Answered question

2022-02-27

Show:

\int_{\frac{1}{3}}^{\frac{1}{2}} \frac{arctanh (t)}{t} d t = \int_{\ln 2}^{\ln 3} \frac{u}{2 sinh u} d u

besplodnexkj

Beginner2022-02-28Added 7 answers

The limits of integration provide a natural starting point:

\ln \frac{1}{2} = - \ln 2, \ln \frac{1}{3} = - \ln 3

, and

- \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{\ln v}{1 - v^{2}} d v = \int_{\frac{1}{2}}^{\frac{1}{3}} \frac{\ln v}{1 - v^{2}} d v

so one should consider the possibility that

u = - \ln v

. If so,

d u = - \frac{1}{v} d v

, and

sinh u = \frac{1}{2} (e^{u} - e^{- u}) = \frac{1}{2} (e^{- \ln v} - e^{\ln v}) = \frac{1}{2} (\frac{1}{v} - v)

= \frac{1 - v^{2}}{2 v}

\int_{\ln 2}^{\ln 3} \frac{u}{2 sinh u} d u = \int_{\frac{1}{2}}^{\frac{1}{3}} \frac{- \ln v}{\frac{1 - v^{2}}{v}} (- \frac{1}{v}) d v = \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{\ln v}{1 - v^{2}} d v

= \int_{\frac{1}{2}}^{\frac{1}{3}} \frac{\ln v}{1 - v^{2}} d v

The other equalities will also succumb to reasonable substitutions.

Tate Puckett

Beginner2022-03-01Added 6 answers

I'd like to give a suggestion for the equality

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

The idea is to rewrite

arctanh t = \int_{0}^{t} \frac{1}{1 - s^{2}} d s

, so that we have

\int_{\frac{1}{3}}^{\frac{1}{2}} \frac{arctanh t}{t} d t = \int_{\frac{1}{3}}^{\frac{1}{2}} \int_{0}^{t} \frac{1}{1 - s^{2}} \frac{1}{t} d s d t

Now we can change the order of integration:

\int_{\frac{1}{3}}^{\frac{1}{2}} \int_{0}^{t} \frac{1}{1 - s^{2}} \frac{1}{t} d s d t = \int_{0}^{\frac{1}{3}} \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{1}{1 - s^{2}} \frac{1}{t} d t d s + \int_{\frac{1}{3}}^{\frac{1}{2}} \int_{s}^{\frac{1}{2}} \frac{1}{1 - s^{2}} \frac{1}{t} d t d s

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

= \frac{1}{2} \log (\frac{3}{2}) \log (2) + \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{\log (\frac{1}{2}) - \log s}{1 - s^{2}} d s

= \frac{1}{2} \log (\frac{3}{2}) \log (2) + \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{\log (\frac{1}{2})}{1 - s^{2}} d s - \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{\log s}{1 - s^{2}} d s

This is helpful 0 Flag

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?