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$$\int_0^\infty \frac{\ln(1+u)}{u+u^\phi}\mathrm du$$

I saw this problem posted before, however it was deleted. I forgot the solution to this integral but I don't know how to begin. Maybe I should try expanding $\ln(1+u)$ but I would be forced to split the integral from 0 to 1 and 1 to $\infty$ with $u \to 1/u$.

It would be easier if the $u$'s on the bottom were in brackets, cuz then I could use the beta function. I tried Feynman's trick with ln(1+au) but that didn't help.

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  • $\begingroup$ Feynman trick as you mentioned turns this into an integral of a rational function which should be much more tractable, e.g with a keyhole contour $\endgroup$
    – K.defaoite
    Commented May 20 at 17:19
  • $\begingroup$ Perhaps my cowardice is my downfall, Ill see what i can do $\endgroup$
    – uggupuggu
    Commented May 20 at 17:22
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    $\begingroup$ A close miss is on page 541 of Table of Integrals, Series, and Products by Gradshteyn and Ryzhik $$\int_0^\infty \frac{x^{p-1}\ln x}{1+x^q}=-\frac{\pi^2}{q^2}\frac{\cos\left(\frac{p\pi}{q}\right)}{\sin\left(\frac{p\pi}{q}\right)^2} \\ p,q\in\mathbb R_+~~,~~p<q$$ However the expression is unfortunately invalid when $p=0$ (our case). Are you sure the integral in question is convergent? $\endgroup$
    – K.defaoite
    Commented May 20 at 17:47
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    $\begingroup$ @K.defaoite But I can $\endgroup$
    – Martin.s
    Commented May 21 at 4:12
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    $\begingroup$ Well done martin $\endgroup$
    – K.defaoite
    Commented May 21 at 14:28

3 Answers 3

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\begin{eqnarray*} \int_0^\infty \frac{\ln(1+u)}{u+u^\phi}du &=& \int_0^1 \frac{\ln(1+u)}{u(1+u^{\phi-1})}du + \int_1^\infty \frac{\ln(1+u)}{u(1+u^{\phi-1})}du \\ &=& \int_0^1 \frac{\ln(1+u)}{u(1+u^{\phi-1})}du + \underbrace{\int_0^1 \frac{\ln\left(\frac{1+u}{u}\right)u^{\phi-1}}{u(1+u^{\phi-1})}du}_{u\mapsto \frac{1}{u}}\\ &=& \int_0^1 \frac{\ln(1+u)}{u(1+u^{\phi-1})}du+ \int_0^1 \frac{\ln(1+u)u^{\phi-1}}{u(1+u^{\phi-1})}du - \int_0^1 \frac{\ln(u)u^{\phi-1}}{u(1+u^{\phi-1})}du\\ &=& \int_0^1 \left[\frac{\ln(1+u)}{u(1+u^{\phi-1})}+ \frac{\ln(1+u)u^{\phi-1}}{u(1+u^{\phi-1})} \right]du - \int_0^1 \frac{\ln(u)u^{\phi-1}}{u(1+u^{\phi-1})}du\\ &=& \int_0^1 \frac{\ln(1+u)}{u}du - \int_0^1 \frac{\ln(u)u^{\phi-1}}{u(1+u^{\phi-1})}du\\ &=& \frac{\pi^2}{12}- \sum_{j=0}^{\infty}(-1)^j\int_0^1 \ln(u)u^{\left[j(\phi-1)+\phi-2\right]}\\ &=& \frac{\pi^2}{12}- \lim_{w \to 0+} \frac{d}{dw}\sum_{j=0}^{\infty}(-1)^j\int_0^1 u^wu^{\left[j(\phi-1)+\phi-2\right]}\\ &=& \frac{\pi^2}{12}- \lim_{w \to 0+} \frac{d}{dw}\sum_{j=0}^{\infty}(-1)^j\int_0^1 u^{\left[j(\phi-1)+\phi-2+w\right]}\\ &=& \frac{\pi^2}{12}- \lim_{w \to 0+} \frac{d}{dw}\sum_{j=0}^{\infty}(-1)^j\frac{1}{(j(\phi-1)+\phi-1+w)}\\ &=& \frac{\pi^2}{12}+ \sum_{j=0}^{\infty}(-1)^j\frac{1}{(j(\phi-1)+\phi-1)^2}\\ &=& \frac{\pi^2}{12}+ \frac{1}{(\phi-1)^2}\sum_{j=0}^{\infty}\frac{(-1)^j}{(j+1)^2} \quad \phi\neq 1\\ &=& \frac{\pi^2}{12}\left[1+\frac{1}{(\phi-1)^2}\right] \end{eqnarray*}

$$\boxed{\int_0^\infty \frac{\ln(1+u)}{u+u^\phi}du = \frac{\pi^2}{12}\left[1+\frac{1}{(\phi-1)^2}\right], \phi > 1 } $$

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    $\begingroup$ Stellar solution $\endgroup$ Commented May 22 at 4:41
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$$ \begin{aligned} & \Omega_{a \in C(1)} = \int_0^{\infty} \frac{\ln (1+u)}{u+u^a} d u = \int_0^1 \frac{\ln (1+u)}{u+u^a} d u + \int_1^{\infty} \frac{\ln (1+u)}{u+u^a} d u \\ & = \int_0^1 \frac{\ln (1+u)}{u+u^a} d u + \int_0^1 \frac{\ln \left(\frac{1+u}{u}\right)}{\frac{1}{u}+\frac{1}{u^a}} \frac{d u}{u^2} \\ & = \int_0^1 \frac{\ln (1+u)}{u+u^a} d u + \int_0^1 \frac{u^a \ln (1+u) - u^a \ln (u)}{u\left(u+u^a\right)} d u \\ & = \int_0^1 \frac{\ln (1+u)}{u+u^a} d u + \int_0^1 \frac{u^a \ln (1+u)}{u\left(u+u^a\right)} d u - \int_0^1 \frac{u^a \ln (u)}{u\left(u+u^a\right)} d u \\ & = \int_0^1 \frac{\left(u+u^a\right) \ln (1+u)}{u\left(u+u^a\right)} d u - \frac{1}{a-1} \int_0^1 \frac{u^{a-2} \ln \left(u^{a-1}\right)}{1+u^{a-1}} d u \\ & = \int_0^1 \frac{\ln (1+u)}{u} d u - \frac{1}{(a-1)^2} \int_0^1 \frac{\ln \left(u^{a-1}\right)}{1+u^{a-1}} d\left(u^{a-1}\right) \end{aligned} $$

$$ \begin{aligned} & = -\left. L i_2(-x) \right|_0^1 - \frac{1}{(a-1)^2} \underbrace{\frac{\ln u}{1+u} d u}_{\underbrace{}_{I B P}} \\ & = - L i_2(-1) - \left. \frac{1}{(a-1)^2} \left\{\ln (x) \ln (1+x) + L i_2(-x)\right\} \right|_0^1 \\ & = - L i_2(-1) - \frac{L i_2(-1)}{(a-1)^2} \\ & = -\frac{(a-1)^2+1}{(a-1)^2} L i_2(-1) \\ & = \frac{(a-1)^2+1}{(a-1)^2} \frac{\zeta(2)}{2} \\ & \text{As } a = \phi: \Omega_a = \int_0^{\infty} \frac{\ln (1+u)}{u+u^\phi} d u \\ & = \frac{\left(\frac{1+\sqrt{5}}{2}-1\right)^2+1}{\left(\frac{1+\sqrt{5}}{2}-1\right)^2} \frac{\zeta(2)}{2} \\ & = \frac{5-\sqrt{5}}{3-\sqrt{5}} \frac{\zeta(2)}{2} \\ & = \frac{(5-\sqrt{5})(3+\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})} \frac{\zeta(2)}{2} \\ & = \frac{1+\sqrt{5}}{2} \cdot \sqrt{5} \cdot \frac{\zeta(2)}{2} \\ & = \frac{\phi \sqrt{5} \zeta(2)}{2} \end{aligned} $$

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  • $\begingroup$ This is incorrect. $\endgroup$
    – K.defaoite
    Commented May 21 at 14:33
  • $\begingroup$ @K.defaoite How so $\endgroup$
    – uggupuggu
    Commented May 21 at 14:44
  • $\begingroup$ @uggupuggu It gives incorrect numerical results for, e.g, $\phi=2$. Click the link. $\endgroup$
    – K.defaoite
    Commented May 21 at 14:55
  • $\begingroup$ Oh my mistake, I see what happened. The poster substituted in $\phi=\frac{1+\sqrt 5}{2}$. I assumed $\phi$ was supposed to be arbitrary. $\endgroup$
    – K.defaoite
    Commented May 21 at 14:57
  • $\begingroup$ @K.defaoite its ok sugar $\endgroup$
    – uggupuggu
    Commented May 21 at 15:00
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For any $\phi>1$, we first split the integral into 2 and then express them as Dilogarithm functions .$$ \begin{aligned} \int_0^{\infty} \frac{\ln (1+u)}{u+u^\phi} d u & =\int_0^1 \frac{\ln (1+u)}{u+u^\phi} d u+\int_1^{\infty} \frac{\ln (1+u)}{u+u^\phi} d u \\ & =\int_0^1 \frac{\ln (1+u)}{u+u^\phi} d u+\int_0^1 \frac{\ln \left(1+\frac{1}{u}\right)}{\frac{1}{u}+\frac{1}{u^\phi}} \frac{d u}{u^2} \\ & =\int_0^1 \frac{\ln (1+u)}{u+u^\phi} d u+\int_0^1 \frac{u^{\phi-1}(\ln (1+u)-\ln u)}{u+u^\phi} d u \\ & =\int_0^1 \frac{\left(1+u^{\phi-1}\right) \ln (1+u)}{u+u^\phi} d u-\int_0^1 \frac{u^{\phi-1} \ln u}{u+u^\phi} d u \\ & =\int_0^1 \frac{\ln (1+u)}{u} d u-\int_0^1 \frac{u^{\phi-2}}{1+u^{\phi-1}} \ln u d u \\ & =-\operatorname{Li_2}(-1)-\frac{1}{\phi-1} \int_0^1 \ln u \, d\left(\ln \left(1+u^{\phi-1}\right)\right) \\ & =\frac{\pi^2}{12}-\frac{1}{\phi-1}\left(\left[\ln u \ln \left(1+u^{\phi-1}\right)\right]_0^1-\int_0^1 \frac{\ln \left(1+u^{\phi-1}\right)}{u}du\right)\\ & =\frac{\pi^2}{12}+\frac{1}{(\phi-1)^2} \int_0^1 \frac{\ln \left(1+u^{\phi-1}\right)}{u^{\phi-1}}\, d\left(u^{\phi-1}\right) \\ & =\frac{\pi^2}{12}\left[1+\frac{1}{(\phi-1)^2}\right] \end{aligned} $$ functions.

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