I'm solving an exercise in complex analysis.

I proved $\displaystyle \gamma=\int_0^1 -\log(\log(1/s)) \, ds$.

I want to prove that $\displaystyle \gamma=\int_0^1 \frac{1-e^{-t}-e^{-1/t}} t \,dt$.

But I don't have any idea. Anyone can help me?

  • 1
    $\begingroup$ Substitute $1/s = e^u$. Looking where you want to end up, see whether you can figure out how to continue. $\endgroup$ – Daniel Fischer Dec 16 '16 at 15:12
  • $\begingroup$ math.stackexchange.com/questions/980593/… $\endgroup$ – tired Dec 16 '16 at 15:13
  • $\begingroup$ It is known as Barne's formula. $\endgroup$ – Jean Marie Dec 16 '16 at 18:37

One way to show two integrals are equal is to transform one into the other. If we start with the first integral, we want to get rid of the logarithms. One logarithm is removed by the substitution $s = e^{-u}$, which gives us

$$\gamma = \int_0^1 -\log \bigl(\log (1/s)\bigr)\,ds = -\int_0^{\infty} e^{-u}\log u\,du.$$

Our target is an integral over $[0,1]$, and we want something with $e^{-1/t}$ in it. That suggests splitting the integral into two parts, and in the integral over $[1,\infty)$ making the substitution $t = 1/u$. But we also want to get rid of the $\log u$, and instead want a factor $\frac{1}{t} = \frac{d}{dt}\log t$. That suggests integrating by parts. So we continue

\begin{align} -\int_0^{\infty} e^{-u}\log u\,du &= -\int_0^1 e^{-u}\log u\,du - \int_1^{\infty} e^{-u}\log u\,du \\ &= -\int_0^1 e^{-t}\log t\,dt + \int_0^1 t^{-2}e^{-1/t}\log t\,dt \\ &= -\int_0^1 \log t \frac{d}{dt}\bigl(1 - e^{-t}\bigr)\,dt + \int_0^1 \log t \frac{d}{dt}\bigl(e^{-1/t}\bigr)\,dt \\ &= -\log t\bigl(1-e^{-t}\bigr)\biggr\rvert_0^1 + \int_0^1 \frac{1-e^{-t}}{t}\,dt + \log t(e^{-1/t})\biggr\rvert_0^1 - \int_0^1 \frac{e^{-1/t}}{t}\,dt \\ &= \int_0^1 \frac{1 - e^{-t} - e^{-1/t}}{t}\,dt \end{align}

and arrive at our goal.


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