Question Number 72391 by mathmax by abdo last updated on 28/Oct/19
![calculte ∫_0 ^∞ (((−1)^([x]) )/(4+x^2 ))dx](Q72391.png)
$${calculte}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{\left[{x}\right]} }{\mathrm{4}+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 01/Nov/19
![∫_0 ^∞ (((−1)^([x]) )/(x^2 +4))dx =Σ_(n=0) ^∞ ∫_n ^(n+1) (((−1)^n )/(x^2 +4))dx =Σ_(n=0) ^∞ (−1)^n ∫_n ^(n+1) (dx/(x^2 +4)) but ∫_n ^(n+1) (dx/(x^2 +4)) =_(x=2t) ∫_(n/2) ^((n+1)/2) ((2dt)/(4t^2 +4)) =(1/2) ∫_(n/2) ^((n+1)/2) (dt/(t^2 +1)) =(1/2)[arctant]_(n/2) ^((n+1)/2) =(1/2){ arctan(((n+1)/2))−arctan((n/2))} ⇒∫_0 ^∞ (((−1)^([x]) )/(x^2 +4))dx =(1/2)Σ_(n=0) ^∞ (−1)^n {arctan(((n+1)/2))−arctan((n/2))} =(1/2)Σ_(n=0) ^∞ (−1)^n arctan(((n+1)/2))−(1/2)Σ_(n=0) ^∞ (−1)^n arctan((n/2)) ...be continued...](Q72756.png)
$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{\left[{x}\right]} }{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{{n}} ^{{n}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \int_{{n}} ^{{n}+\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:{but}\:\int_{{n}} ^{{n}+\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:=_{{x}=\mathrm{2}{t}} \:\:\:\int_{\frac{{n}}{\mathrm{2}}} ^{\frac{{n}+\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{2}{dt}}{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\frac{{n}}{\mathrm{2}}} ^{\frac{{n}+\mathrm{1}}{\mathrm{2}}} \:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}}\left[{arctant}\right]_{\frac{{n}}{\mathrm{2}}} ^{\frac{{n}+\mathrm{1}}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{arctan}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)−{arctan}\left(\frac{{n}}{\mathrm{2}}\right)\right\} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\left[{x}\right]} }{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \left\{{arctan}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)−{arctan}\left(\frac{{n}}{\mathrm{2}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{arctan}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{arctan}\left(\frac{{n}}{\mathrm{2}}\right) \\ $$$$...{be}\:{continued}... \\ $$