Question Number 91619 by abdomathmax last updated on 02/May/20
![calculate ∫_2 ^(+∞) (((−1)^([2x]) )/(x[x]−1))dx](https://www.tinkutara.com/question/Q91619.png)
$${calculate}\:\int_{\mathrm{2}} ^{+\infty} \:\frac{\left(−\mathrm{1}\right)^{\left[\mathrm{2}{x}\right]} }{{x}\left[{x}\right]−\mathrm{1}}{dx} \\ $$
Commented by mathmax by abdo last updated on 02/May/20
![=Σ_(n=2) ^∞ ∫_n ^(n+1) (((−1)^([2x]) )/(nx−1))dx n<x<n+1 ⇒2n<2x<2n+2 ⇒2x ∈[2n,2n+1]∪[2n+1,2n+2[ =Σ_(n=2) ^∞ ∫_n ^(n+(1/2)) (((−1)^(2n) )/(nx−1)) +Σ_(n=2) ^∞ ∫_(n+(1/2)) ^(n+1) (((−1)^(2n+1) )/(nx−1))dx =Σ_(n=2) ^∞ (1/n)[ln(nx−1)]_n ^(n+(1/2)) −Σ_(n=2) ^∞ (1/n)[ln(nx−1)]_(n+(1/2)) ^(n+1) =Σ_(n=2) ^∞ (1/n){ln(n^2 +(n/2)−1)−ln(n^2 −1))} −Σ_(n=2) ^∞ (1/n){ln(n^2 +n−1)−ln(n^2 +(n/2)−1)} =Σ_(n=2) ^∞ (1/n)ln(((n^2 +(n/2)−1)/(n^2 −1)))−Σ_(n=2) ^∞ (1/n)ln(((n^2 +n−1)/(n^2 +(n/2)−1)))} ...be continued...](https://www.tinkutara.com/question/Q91706.png)
$$=\sum_{{n}=\mathrm{2}} ^{\infty} \:\int_{{n}} ^{{n}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\left[\mathrm{2}{x}\right]} }{{nx}−\mathrm{1}}{dx} \\ $$$${n}<{x}<{n}+\mathrm{1}\:\Rightarrow\mathrm{2}{n}<\mathrm{2}{x}<\mathrm{2}{n}+\mathrm{2}\:\Rightarrow\mathrm{2}{x}\:\in\left[\mathrm{2}{n},\mathrm{2}{n}+\mathrm{1}\right]\cup\left[\mathrm{2}{n}+\mathrm{1},\mathrm{2}{n}+\mathrm{2}\left[\right.\right. \\ $$$$=\sum_{{n}=\mathrm{2}} ^{\infty} \:\int_{{n}} ^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{n}} }{{nx}−\mathrm{1}}\:+\sum_{{n}=\mathrm{2}} ^{\infty} \:\int_{{n}+\frac{\mathrm{1}}{\mathrm{2}}} ^{{n}+\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }{{nx}−\mathrm{1}}{dx} \\ $$$$=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\left[{ln}\left({nx}−\mathrm{1}\right)\right]_{{n}} ^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} \:−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\left[{ln}\left({nx}−\mathrm{1}\right)\right]_{{n}+\frac{\mathrm{1}}{\mathrm{2}}} ^{{n}+\mathrm{1}} \\ $$$$\left.=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\left\{{ln}\left({n}^{\mathrm{2}} \:+\frac{{n}}{\mathrm{2}}−\mathrm{1}\right)−{ln}\left({n}^{\mathrm{2}} −\mathrm{1}\right)\right)\right\} \\ $$$$−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\left\{{ln}\left({n}^{\mathrm{2}} +{n}−\mathrm{1}\right)−{ln}\left({n}^{\mathrm{2}} \:+\frac{{n}}{\mathrm{2}}−\mathrm{1}\right)\right\} \\ $$$$\left.=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}}{ln}\left(\frac{{n}^{\mathrm{2}} \:+\frac{{n}}{\mathrm{2}}−\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}}\right)−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}}{ln}\left(\frac{{n}^{\mathrm{2}} \:+{n}−\mathrm{1}}{{n}^{\mathrm{2}} \:+\frac{{n}}{\mathrm{2}}−\mathrm{1}}\right)\right\} \\ $$$$…{be}\:{continued}… \\ $$