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Prove that

$$\int_{0}^{1}{x+x^2+x^3+\cdots+x^{2n}-2nx\over (1+x)\ln{x}}dx=\ln{\left[\left({2\over \pi}\right)^n(2n)!!\right]}$$

$n\ge1$

My try:

$${x+x^2+\cdots+x^{2n}\over 1+x}={1\over 1+x}\cdot{x(1-x^{2n})\over 1-x}=x\cdot{1-x^{2n}\over 1-x^2}$$

$${x+x^2+\cdots+x^{2n}-2nx\over 1+x}=x\cdot{1-x^{2n}\over 1-x^2}-{2nx\over 1+x}$$

$$\int_{0}^{1}\left(x\cdot{1-x^{2n}\over 1-x^2}-{2nx\over 1+x}\right)\cdot{1\over \ln{x}}dx$$

I am stuck not quite sure what to do next, please help

Edit

$$\int_{0}^{1}\left(x\cdot{1-x^{2n}\over 1-x^2}-{2nx\over 1+x}\right)\cdot{1\over \ln{x}}dx$$

Blencer
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gymbvghjkgkjkhgfkl
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3 Answers3

5

Hint. One may use the identity

$$ \int_0^\infty \frac{1-e^{-su}}{u(e^u+1)}du=\frac12\log(\pi)+\log \Gamma\left(1+\frac{s}2\right)-\log \Gamma\left(\frac{1+s}2\right), \quad s>0. $$

proved here. By the change of variable $u=-\ln x$, $x=e^{-u}$, in the preceding integral, one gets $$ \int_{0}^{1}{x^s-1\over (1+x)\ln{x}}\:dx=\log\left(\frac{\sqrt{\pi}\:\Gamma\left(1+\frac{s}2\right)}{\Gamma(\frac{1+s}2)} \right), \quad s>0, $$ then rearranging the numerator of the given integral as $$ \int_{0}^{1}{(x^2-x)+(x^3-x)+\cdots+(x^{2n}-x)\over (1+x)\ln{x}}\:dx, $$ observing that each term $x^k-x$ rewrites $(x^k-1)-(x-1)$ yields the announced result.

Community
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Olivier Oloa
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4

In THIS ANSWER, I showed that

$$\bbox[5px,border:2px solid #C0A000]{\int_0^1 \left(\frac{x^{2n}-x}{1+x}\right)\,\frac{1}{\log(x)}\,dx=\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right)}$$

We can use this result to evaluate the integral

$$\begin{align} \int_0^1 \left(\frac{x^{2n+1}-x}{1+x}\right)\,\frac{1}{\log(x)}\,dx&=\int_0^1 \left(\frac{x^{2n+1}-x^2+(x^2-x)}{1+x}\right)\,\frac{1}{\log(x)}\,dx\\\\ &=\int_0^1 \left(\frac{x^{2n+1}-x^2}{1+x}\right)\,\frac{1}{\log(x)}\,dx+\log(4/\pi)\\\\ &=\int_0^1 \frac{x^{2n}-x}{\log(x)}\,dx+\log(4/\pi)-\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right)\tag1 \end{align}$$

The integral on the right-hand side of $(1)$ can be evaluated by enforcing the substitution $x\to e^{-x}$ and evaluating the resulting FRULLANI INTEGRAL. Proceeding, we have

$$\int_0^1 \frac{x^{2n}-x}{\log(x)}\,dx=\log\left(\frac{2n+1}{2}\right) \tag 2$$

Substituting $(2)$ into $(1)$ reveals

$$\begin{align} \int_0^1 \left(\frac{x^{2n+1}-x}{1+x}\right)\,\frac{1}{\log(x)}\,dx&=\log\left(\frac{2n+1}{2}\right)+\log(4/\pi)-\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right)\\\\ &= \log\left(\frac{2}{\pi}(2n+1)\right)-\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right) \tag 3\\\\ &=\log\left(\frac{(2n+1)!!}{(2n)!!}\right) \end{align}$$

Finally, we see that

$$\begin{align} \int_0^1 \left(\frac{x+x^2+\cdots +x^{2n}-2nx}{1+x}\right)\,\frac{1}{\log(x)}\,dx&=\sum_{k=1}^n \int_0^1 \frac{x^{2k}-x}{1+x}\frac{1}{\log(x)}\,dx\\\\ &+\sum_{k=1}^{n-1}\int_0^1 \frac{x^{2k+1}-x}{1+x}\frac{1}{\log(x)}\,dx\\\\ &=\sum_{k=1}^n\log\left(\frac{2}{\pi}\frac{(2k)!!}{(2k-1)!!}\right)\\\\ &+\sum_{k=1}^{n-1}\log\left(\frac{2}{\pi}(2n+1)\right)\\\\ &-\sum_{k=1}^n\log\left(\frac{2}{\pi}\frac{(2k)!!}{(2k-1)!!}\right)\\\\ &=\log\left(\left(\frac{2}{\pi}\right)^n (2n)!!\right) \end{align}$$

as was to be shown!

Community
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Mark Viola
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2

Hint: We may use the substitution $x\to e^{-x};$ $$I(n) = \int_0^\infty \left( \frac{e^{-2(n+1)t}-e^{-2t}}{1-e^{-2t}} +\frac{2ne^{-2t}}{1+e^{-t}}\right)t^{-1}dt\\ = \sum_{k=0}^\infty \int_0^\infty\left(e^{-2(n+k+1)t} - e^{-2(k+1)t} +2n (-1)^k e^{-(k+2)t}\right) t^{-1}dt$$ and write the previous splitting in terms of $\Gamma-$function

A. PI
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