Question Number 205173 by mr W last updated on 12/Mar/24
$${find} \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$
Commented by lepuissantcedricjunior last updated on 12/Mar/24
![S=Σ_(n=1) ^∞ (1/((a^2 +n^2 )^2 ))=∫_0 ^∞ (dx/((a^2 +x^2 )^2 )) S=∫_0 ^(+∞) (dx/(a^4 [1+((x/a))^2 ]^2 )) posons (x/a)=tan𝛃⇔dx=a(1+tan^2 𝛃)d𝛃 qd: { ((x→0)),((x→∞)) :}⇒ { ((𝛃→0)),((𝛃→(𝛑/2))) :} S=∫_0 ^(𝛑/2) (d𝛃/(a^3 (1+tan^2 𝛃)))=∫_0 ^(𝛑/2) ((cos^2 𝛃)/a^3 )d𝛃 S=(1/a^3 )[(1/2)𝛃+(1/4)sin2𝛃]_0 ^(𝛑/2) S=(𝛑/(4a^3 ))](https://www.tinkutara.com/question/Q205178.png)
$$\boldsymbol{{S}}=\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{n}}^{\mathrm{2}} \right)^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{dx}}}{\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{x}}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\boldsymbol{{S}}=\int_{\mathrm{0}} ^{+\infty} \frac{\boldsymbol{{dx}}}{\boldsymbol{{a}}^{\mathrm{4}} \left[\mathrm{1}+\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{a}}}\right)^{\mathrm{2}} \right]^{\mathrm{2}} } \\ $$$$\boldsymbol{{posons}}\:\frac{\boldsymbol{{x}}}{\boldsymbol{{a}}}=\boldsymbol{{tan}\beta}\Leftrightarrow\boldsymbol{{dx}}=\boldsymbol{{a}}\left(\mathrm{1}+\boldsymbol{{ta}}\overset{\mathrm{2}} {\boldsymbol{{n}}\beta}\right)\boldsymbol{{d}\beta} \\ $$$$\boldsymbol{{qd}}:\begin{cases}{\boldsymbol{{x}}\rightarrow\mathrm{0}}\\{\boldsymbol{{x}}\rightarrow\infty}\end{cases}\Rightarrow\begin{cases}{\boldsymbol{\beta}\rightarrow\mathrm{0}}\\{\boldsymbol{\beta}\rightarrow\frac{\boldsymbol{\pi}}{\mathrm{2}}}\end{cases} \\ $$$$\boldsymbol{{S}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\boldsymbol{{d}\beta}}{\boldsymbol{{a}}^{\mathrm{3}} \left(\mathrm{1}+\boldsymbol{{ta}}\overset{\mathrm{2}} {\boldsymbol{{n}}\beta}\right)}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\boldsymbol{{co}}\overset{\mathrm{2}} {\boldsymbol{{s}}\beta}}{\boldsymbol{{a}}^{\mathrm{3}} }\boldsymbol{{d}\beta} \\ $$$$\boldsymbol{\mathrm{S}}=\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{3}} }\left[\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\beta}+\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{sin}}\mathrm{2}\boldsymbol{\beta}\right]_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \\ $$$$\boldsymbol{\mathrm{S}}=\frac{\boldsymbol{\pi}}{\mathrm{4}\boldsymbol{{a}}^{\mathrm{3}} } \\ $$$$ \\ $$
Commented by mr W last updated on 12/Mar/24
$${how}\:{can}\:{you}\:{get} \\ $$$$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{n}}^{\mathrm{2}} \right)^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{dx}}}{\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{x}}^{\mathrm{2}} \right)^{\mathrm{2}} }\:? \\ $$
Answered by Berbere last updated on 13/Mar/24
$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)}={f}\left({a}\right) \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}{ai}}\left(\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}−{ia}}−\frac{\mathrm{1}}{{n}+{ia}}\right)=\frac{\mathrm{1}}{\mathrm{2}{ai}}\left(\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}−{ia}} −{t}^{{n}+{ia}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ai}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{−{ia}} −{t}^{{ia}} }{\mathrm{1}−{t}}{dt};\Psi\left({z}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{z}−\mathrm{1}} }{\mathrm{1}−{x}}{dx} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}{ai}}\left(\Psi\left(\mathrm{1}+{ia}\right)−\Psi\left(\mathrm{1}−{ia}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ai}}\left(\frac{\mathrm{1}}{{ia}}+\Psi\left({ia}\right)−\Psi\left(\mathrm{1}−{ia}\right)\right) \\ $$$$\Psi\left(\mathrm{1}−{z}\right)−\Psi\left({z}\right)=\pi{cot}\left(\pi{z}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}{ai}}\pi{cot}\left({i}\pi{a}\right)=−\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }+\frac{\pi}{\mathrm{2}{a}}{coth}\left(\pi{a}\right) \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}\leqslant\zeta\left(\mathrm{2}\right)\:\:\:{serie}\:{Cv}\:{unifomly}\:\Rightarrow \\ $$$${a}\overset{{f}} {\rightarrow}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}\:{is}\:{derivable}?{we}\:{can}\:{change}\:\Sigma\:{and}\:{derivation} \\ $$$${f}'\left({a}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{−\mathrm{2}{a}}{\left({n}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}^{\mathrm{3}} }−\frac{\pi}{\mathrm{2}{a}^{\mathrm{2}} }{coth}\left(\pi{a}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} }.\frac{\mathrm{1}}{{sinh}^{\mathrm{2}} \left({a}\right)} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{4}} }+\frac{\pi}{\mathrm{4}{a}^{\mathrm{3}} }{coth}\left(\pi{a}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} {sinh}^{\mathrm{2}} \left(\pi{a}\right)} \\ $$$$ \\ $$
Commented by mr W last updated on 13/Mar/24
$${thanks}\:{sir}! \\ $$$${nice}\:{solution}! \\ $$
Commented by Berbere last updated on 13/Mar/24
$${Yes}\:{sir}\:\:{Withe}\:{pleasur}\: \\ $$$${i}\:{lost}\:{my}\:{Accunt}\:{Witcher}\: \\ $$$${God}\:{bless}\:{you} \\ $$
Commented by mr W last updated on 13/Mar/24
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