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A friend of mine sent me an integral that she had not been able to crack, and me neither. It comes with a result, but without a proof (I suppose it originated in some math contest). Could you please suggest an approach to prove the result?

$$\int_0^1\frac{x^9\left(x^4+x^2-x-1-5\ln x\right)}{\left(x^{10}-1\right)\ln x}\mathrm dx=\frac12\gamma+\frac{11}5\ln2-\frac54\ln5+\frac12\ln\pi-\frac12\ln\phi,$$ where $\gamma$ is the Euler–Mascheroni constant, and $\phi$ is the golden ratio.

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    $\begingroup$ @Alizter: You did not just bump a bunch of old, well voted and well received threads just to romanized the d in the integral signs, right? Please tell me you didn't. (And if you did, please let it go and don't do that again. This is forcing your style choices on other people; if it's not $\LaTeX$'d at all, fine, but if it's already reasonably readable output... you should leave it be.) $\endgroup$ – Asaf Karagila Aug 29 '14 at 23:18
  • $\begingroup$ @AsafKaragila Yes I unfortunately did this. I realise now it was wrong. I am not thinking sorry. $\endgroup$ – Ali Caglayan Aug 29 '14 at 23:23
  • $\begingroup$ @Alizter: As long as you understand the mistake, it's fine. $\endgroup$ – Asaf Karagila Aug 29 '14 at 23:28
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Let's denote the integral in question as $$I=\int_0^1\frac{x^9\left(x^4+x^2-x-1-5\ln x\right)}{\left(x^{10}-1\right)\ln x}dx.\tag1$$ Changing the variable $x=y^{1/10}$ and renaming $y$ back to $x$ we get $$I=\int_0^1\frac{x^{2/5}+x^{1/5}-x^{1/10}-1-\ln\sqrt x}{(x-1)\ln x}dx.\tag2$$ Some elementary transformations show that $$I=\mathcal{J}(2/5)+\mathcal{J}(1/5)-\mathcal{J}(1/10),\tag3$$ where we introduced notation $$\mathcal{J}(q)=\int_0^1\frac{x^q-1-q\,\ln x}{(x-1)\ln x}dx.\tag4$$ The integral $\mathcal{J}(q)$ can be evaluated as follows: $$\begin{align}\mathcal{J}(q)=\int_0^1\int_0^q\frac{x^p-1}{x-1}dp\,dx=\int_0^q\underbrace{\int_0^1\frac{x^p-1}{x-1}dx}_{\text{DLMF 5.9.16}}\,dp\\=\int_0^q H_p\,dp=q\cdot\gamma+\ln\Gamma(q+1),\end{align}\tag5$$ where $H_p$ are harmonic numbers: $H_p$$\,=\,$$\gamma$$\,+\,$$\psi_0$$(p+1)$, and $\psi_0$ is the digamma function: $\psi_0(x)=\frac{d}{dx}\ln\,$$\Gamma$$(x)$. Let me mention that the formula DLMF 5.9.16 becomes particularly obvious for positive integer $p$, when $H_p=\sum_{n=1}^pn^{-1}$.

Pluging $(5)$ back into $(3)$, we get $$I=\frac12\gamma+\ln\frac45+\ln\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac25\right)}{\Gamma\left(\frac1{10}\right)}.\tag6$$ From the formula $(74)$ on this MathWorld page we know that $$\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac25\right)}{\Gamma\left(\frac1{10}\right)}=\frac{\sqrt[5]2\,\sqrt\pi}{\sqrt[4]5\,\sqrt\phi}.\tag7$$ (see the paper Raimundas Vidūnas, Expressions for values of the gamma function for a proof).

Making use of this formula, we get the final result $$I=\frac12\gamma+\frac{11}5\ln2-\frac54\ln5+\frac12\ln\pi-\frac12\ln\phi.\tag8$$

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    $\begingroup$ Simply Amazing! $\endgroup$ – nbubis Oct 23 '13 at 21:41
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    $\begingroup$ Really brilliant Vladimir. $\endgroup$ – Bennett Gardiner Oct 24 '13 at 3:31

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