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I have a question which askes us to find the value of the integral:

$\displaystyle \tag*{} \int \limits_{0}^{1}\frac{\ln^2{x}+2\ln{x}-2x+2}{\ln^2{x}-x\ln^2{x}} \mathrm dx$

I tried using differentiation under integral using some variable, I couldn't go any further, I even tried the substitution $\ln x=t$, it just reduces the size of integral. I divided the integral with the denominator to get $3$ separate integrals but each of them diverges. Any help would be appreciated. Thanks.

2 Answers2

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We first begin by enforcing the substitution $e^{-u} = x \, \implies -e^{-u} \, du= dx$

$$I=\int_{0}^{1} \frac{\ln^2 x+2\ln x-2x+2}{\ln^2 x - x \ln^2 x} \, dx \stackrel{x \mapsto e^{-u}}{=}\int_{0}^{\infty} \frac{1}{u^2} \cdot \frac{u^2-2u-2e^{-u} + 2}{e^u-1} \, du$$

Now using the fact that: $$\zeta(s)\Gamma(s)=\int_{0}^{\infty} \frac{x^{s-1}}{e^x-1} \, dx$$ We can then compute the following Mellin transform:

$$\int_{0}^{\infty} x^{s-1} \frac{x^2-2x-2e^{-x}+2}{e^x-1} \, dx=\Gamma (s) \left(2-2s \zeta (1+s) +s (1+s) \zeta (2+s)\right)$$

We are interested in the case that $s\to -1$ which gives: $$\boxed{I=\gamma+\ln (2\pi)-2}$$

KStarGamer
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    Could you please comment on the $s$-region of validity of the various formulae? The one for $\zeta(s)\Gamma(s)$ is for example valid for $\Re s>1$. Also about the interchange of the limit and integration at the end. – Gary Mar 02 '22 at 00:48
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Decompose the integral into three \begin{align} I=&\int_{0}^{1} \frac{\ln^2 x+2\ln x-2x+2}{(1- x) \ln^2 x} \, dx \\ =& \int_0^1 \underset{=\>\ln\frac\pi2}{\frac{x-1}{(x+1) \ln x}} dx + \int_0^1 \underset{=\>\gamma}{\frac{\ln x+1-x}{(1- x) \ln x} }dx -2\int_0^1 \underset{=\>1-\ln2}{\frac{\ln x(1+x^2)+1-x^2}{(1-x^2) \ln^2x} }dx\\ =& \> \ln(2\pi)-2+\gamma \end{align}

Quanto
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