INTEGRALBOT BETA-DIGAMMA INTEGRAL

Today we will show the following result that appears in this twitter post

$\int_{0}^{\infty} \frac{x}{1+x^{3}} \ln \left(1+\frac{1}{x^{3}}\right) d x=\frac{\pi \ln (3)}{\sqrt{3}}-\frac{\pi^{2}}{9}$

We will start by proving two lemmas involving the Beta and Digamma functions.

Lemma 1:

$\int_0^\infty \frac{x^{s-1}\ln(x)}{1+x}\,dx=\Gamma(s)\Gamma(1-s)\left(\psi(s)-\psi(1-s) \right)$ (1)

Recall the Beta function

$I(s)=\int_0^\infty \frac{x^{s-1}}{1+x}\,dx=\Gamma(s)\Gamma(1-s)$ (2)

If we differentiate (2) w.r. to s we obtain

$\begin{aligned} I^\prime(s)&=\int_0^\infty \frac{x^{s-1}\ln(x)}{1+x}\,dx=\frac{d}{ds}\left[\Gamma(s)\Gamma(1-s)\right]\\ &=\Gamma^\prime(s)\Gamma(1-s)-\Gamma(s)\Gamma^\prime(1-s)\\ &=\Gamma(s)\Gamma(1-s)\psi(s)-\Gamma(s)\Gamma(1-s)\psi(1-s)\\ &=\Gamma(s)\Gamma(1-s)\left(\psi(s)-\psi(1-s) \right) \qquad \blacksquare \end{aligned}$

Lemma 2:

$\int_0^1 x^{a-1}(1-x)^{b-1}\ln(x)\,dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\left(\psi(a)-\psi(a+b) \right)$
(3)

Recall the Beta function

$I(a)=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ (4)

If we differentiate (4) w.r. to a we obtain

$\begin{aligned} I^\prime(a)&=\int_0^1 x^{a-1}(1-x)^{b-1}\ln(x)\,dx=\frac{d}{da}\left[\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\right]\\ &=\frac{\Gamma^\prime(a)\Gamma(b)\Gamma(a+b)-\Gamma(a)\Gamma(b)\Gamma^\prime(a+b)}{\Gamma^2(a+b)}\\ &=\frac{\psi(a)\Gamma(a)\Gamma(b)\Gamma(a+b)-\Gamma(a)\Gamma(b)\Gamma(a+b)\psi(a+b)}{\Gamma^2(a+b)}\\ &=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\left(\psi(a)-\psi(a+b) \right) \qquad \blacksquare \end{aligned}$

Now let´s evaluate the integral:

$\begin{aligned} \int_{0}^{\infty} \frac{x}{1+x^{3}} \ln \left(1+\frac{1}{x^{3}}\right) \,d x&=\int_{0}^{\infty} \frac{x}{1+x^{3}} \ln \left(\frac{1+x^{3}}{x^{3}}\right) \,d x\\ &=\int_{0}^{\infty} \frac{x}{1+x^{3}} \ln \left(1+x^{3}\right) \,d x-3\int_{0}^{\infty} \frac{x}{1+x^{3}} \ln \left(x\right) \,d x\\ &=J-3K \end{aligned}$

$\begin{aligned} J&=\int_{0}^{\infty} \frac{x}{1+x^{3}} \ln \left(1+x^{3}\right) \,d x\\ &=\frac13\int_1^\infty \frac{(x-1)^{\frac23-1}\ln(x)}{x}\,dx \qquad \left(1+x^3 \to x \right)\\ &=-\frac13 \int_0^1x^{-\frac23}(1-x)^{\frac23-1}\ln(x),dx\\ &=-\frac13 \frac{\Gamma\left(\frac13\right)\Gamma\left(\frac23\right)}{\Gamma(1)}\left(\psi\left(\frac13\right)-\psi(1) \right) \qquad \left(\text{by lemma 2} \right)\\ &=-\frac{\pi}{3}\frac{1}{\sin\left(\frac{\pi}{3} \right)}\left(-\frac{\pi}{2 \sqrt{3}}-\frac{3 \ln(3)}{2}-\gamma+ \gamma\right)\\ &=\frac{\pi^2}{9}+\frac{\pi \ln(3)}{\sqrt{3}} \qquad \blacksquare \end{aligned}$

For the proof of the special values of the Digamma function see this post.

$\begin{aligned} K&=\int_{0}^{\infty} \frac{x}{1+x^{3}} \ln \left(x\right) \,d x\\ &=\frac13 \int_0^\infty \frac{x^{\frac23-1}\ln(x^{1/3})}{1+x}\,dx \qquad \left(x^3 \to x \right)\\ &=\frac19 \int_0^\infty \frac{x^{\frac23-1}\ln(x^{1/3})}{1+x}\,dx\\ &=\frac19\Gamma\left(\frac23\right)\Gamma\left(\frac13\right)\left(\psi\left(\frac23\right)-\psi\left(\frac13\right) \right)\\ &=-\frac{\pi}{9\sin\left(\frac{2\pi}{3} \right)}\left(\psi\left(\frac13\right)-\psi\left(\frac23\right) \right)\\ &=-\frac{2\pi}{9\sqrt{3}}\left(-\frac{\pi}{\sqrt{3}}\right)\\ &=\frac{2 \pi^2}{27} \qquad \blacksquare \end{aligned}$

Plugging the values of J and K back in the original integral we obtain:

$\begin{aligned} \int_{0}^{\infty} \frac{x}{1+x^{3}} \ln \left(1+\frac{1}{x^{3}}\right) \,d x &=J-3K\\ &=\frac{\pi^2}{9}+\frac{\pi \ln(3)}{\sqrt{3}}-3\frac{2 \pi^2}{27}\\ &=\frac{\pi \ln (3)}{\sqrt{3}}-\frac{\pi^{2}}{9} \qquad \blacksquare \end{aligned}$

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