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How to calculate this integral $$\int_0^1 \frac{1-x}{(x^2-x+1)\log(x)}\;dx$$ In WolframAlpha, I found it equal to $$\log \left[ \frac{\sqrt{\pi}\;\Gamma\left(\frac23\right)}{\Gamma\left(\frac16\right)} \right]$$ I tried using the relation$\quad\int_0^1 x^t\,dt = \frac{x-1}{\log x}\quad$ then dividing both sides by $\quad(x^2-x+1)\quad$ and calculating the double integral $\quad\int_0^1 \int_0^1 \frac{x^t}{x^2-x+1}\,dtdx\quad$ but I'm unable to calculate $$\int_0^1 \frac{x^t}{x^2-x+1}\,dx$$
Any hints?
TIA

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Multiply both the numerator and denominator by $1+x$ so the denominator now turns into the nicely looking $1+x^3$ factor instead of that hideous quadratic factor! Calling the integral $I$ gives$$I\stackrel{\text{def}}{=}\int\limits_0^1dx\,\frac {1-x^2}{\log x(1+x^3)}$$ This resulting integral is actually very trivial when you know how to evaluate it. In fact, the natural logarithm in the denominator reminds us (hopefully) of a very similar integral which can be quickly evaluated using Feynman's Trick$$J(a)=\int\limits_0^1dx\,\frac {x^a-1}{\log x}=\log(a+1)$$ Similarly, in this case, we make the exponent $a$ instead of two to consider the generalized case and differentiate with respect to the parameter. Calling the new integral $I(a)$, not to be confused with $I$, gives$$\begin{align*}I'(a) & =\frac {\partial}{\partial a}\int\limits_0^1dx\,\frac {1-x^a}{\log x(1+x^3)}\\ & =-\int\limits_0^1dx\,\frac {x^a}{1+x^3}\end{align*}$$ Using the infinite geometric series and integrating termwise, we get $$I'(a)=\frac 16\psi_0\left(\frac {a+1}6\right)-\frac 16\psi_0\left(\frac {a+4}6\right)$$ Integrating to return back to our original integral $I(a)$, and recalling that $\psi(z)=\left[\log\Gamma(z)\right]'$, we get$$I(a)=\log\Gamma\left(\frac {a+1}6\right)-\log\Gamma\left(\frac {a+4}6\right)+\log\left[\frac {\Gamma\left(\frac 23\right)}{\Gamma\left(\frac 16\right)}\right]$$ The constant is readily found by substituting $a=0$ into $I(a)$ and observing that our integral evaluates to zero. Then it's just a matter of isolating $C$ and simplifying the result. When $a=2$, we get $$\int\limits_0^1 dx\,\frac {1-x^2}{\log x(x^2-x+1)}\color{blue}{=\log\frac {\sqrt{\pi}\log\left(\frac 23\right)}{\Gamma\left(\frac 16\right)}}$$ Which was to be shown.

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Before you introduce the parameter, multiply the numerator and denominator by $1+x$. The integral you end up with can be expressed in terms of the Digamma Function.

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  • $\begingroup$ thx or do you mean expressed in terms of digamma function? $\endgroup$ – Wolfdale Jun 23 '18 at 2:41
  • $\begingroup$ Yes, of course. My bad. $\endgroup$ – Jack Lam Jun 23 '18 at 5:02

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