Question Number 144500 by mathmax by abdo last updated on 25/Jun/21
$$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} \:+\mathrm{4}} \\ $$
Answered by Dwaipayan Shikari last updated on 26/Jun/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}{a}}{coth}\left(\frac{\pi}{{a}}\right)−\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{4}}=\frac{\pi}{\mathrm{4}}{coth}\left(\frac{\pi}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{4}}=\frac{\pi}{\mathrm{4}}\mathrm{coth}\:\left(\frac{\pi}{\mathrm{2}}\right)+\frac{\mathrm{7}}{\mathrm{8}} \\ $$
Answered by ArielVyny last updated on 25/Jun/21
![we have f(t)=(1/(t^2 +4)) positiv and decreasing in [0.∞] then ∫_0 ^∞ (1/(t^2 +4))dt=∫_0 ^∞ (1/(4(1+(t^2 /4))))dt=(1/4)∫_0 ^∞ (1/(1+(t^2 /4)))dt u=(t/4) du=(1/4)dt ∫_0 ^∞ (1/(1+u^2 ))du=[arctg]_0 ^∞ =(π/2)](https://www.tinkutara.com/question/Q144501.png)
$${we}\:{have}\:{f}\left({t}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{4}}\:{positiv}\:{and}\:{decreasing} \\ $$$${in}\:\left[\mathrm{0}.\infty\right]\:{then} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{4}}{dt}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+\frac{{t}^{\mathrm{2}} }{\mathrm{4}}\right)}{dt}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}{dt} \\ $$$${u}=\frac{{t}}{\mathrm{4}}\:{du}=\frac{\mathrm{1}}{\mathrm{4}}{dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}=\left[{arctg}\right]_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 26/Jun/21
$$\mathrm{the}\:\mathrm{Q}\:\mathrm{is}\:\mathrm{a}\:\mathrm{sum}\:\mathrm{not}\:\mathrm{integral}\:\mathrm{sir}\:\mathrm{viny}… \\ $$