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Calculate $\int_{10}^{100} \frac{\ln(1+x)}{x} dx$ with $10^{-3}$ accuracy.

I don't really see how I should be able to use function series to solve this one, but I'm supposed to according to the text of the exercise.

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    do you know the Taylor expansion for $ln(1+x)$? – 2017-01-04
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    right! this should be enoguh. thanks! – 2017-01-04
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    Start with the change of variable $x\to1/x$ to get the integral to be computed as $I+J$ with $$I=\int_{1/100}^{1/10}\frac{\ln(1+x)}xdx\qquad J=\int_{1/100}^{1/10}\frac{\ln(x)}xdx$$ Compute $I$ by expanding the logarithm (for a precision of $10^{-3}$, three terms are enough) and note that $$2J=\left.(\ln x)^2\right|_{1/100}^{1/10}=-3(\ln 10)^2$$ – 2017-01-04
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    *Please replace $I+J$ by $I-J$ in the coment above. Sorry for the typo. – 2017-01-04

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Through the substitution $z=\frac{1}{x}$, $$ I = \int_{10}^{100}\frac{\log(1+x)}{x}\,dx = \int_{\frac{1}{100}}^{\frac{1}{10}}\frac{\log(z+1)-\log z}{z}\,dx=\left.-\text{Li}_2(-z)-\frac{\log^2(z)}{2}\right|_{\frac{1}{100}}^{\frac{1}{10}}$$ Hence in order to provide an accurate approximation of $I$ is is enough to have accurate approximations of $\text{Li}_2\left(-\frac{1}{10}\right)$, $\text{Li}_2\left(-\frac{1}{100}\right)$ and $\log(10)$. Since $$ \forall z\in(-1,1),\qquad \text{Li}_2(z) = \sum_{n\geq 1}\frac{z^n}{n^2} $$ we have $\text{Li}_2\left(-\frac{1}{10}\right)\approx -\frac{1757}{18000}$ and $\text{Li}_2\left(-\frac{1}{100}\right)\approx -\frac{399}{40000}$.

An accurate approximation of $\log(10)$ can be derived from the Machin-like identity $$ \log(10) = \text{arctanh}\frac{9}{41}+3\,\text{arctanh}\frac{3}{5} $$ and the Taylor expansion for the $\text{arctanh}$ function up to the $x^8$ term. Considering a truncation of the continued fraction of such rational number, we get $\log(10)\approx \frac{19436}{8441}$. Putting everything together, we get:

$$ I \approx\frac{1757}{18000}-\frac{399}{4000}+\frac{3}{2}\left(\frac{19436}{8441}\right)^2 = \color{green}{8.040}38444\ldots$$