Question Number 53783 by maxmathsup by imad last updated on 25/Jan/19
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\mathrm{2}} }{{e}^{{t}} −\mathrm{1}}{dt}\:{interms}\:{of}\:\xi\left(\mathrm{3}\right) \\ $$
Answered by Smail last updated on 26/Jan/19
![A∫_0 ^∞ (t^2 /(e^t −1))dt=∫_0 ^∞ ((t^2 e^(−t) )/(1−e^(−t) ))dt =∫_0 ^∞ t^2 e^(−t) Σ_(n=0) ^∞ e^(−nt) dt =Σ_(n=0) ^∞ ∫_0 ^∞ t^2 e^(−(n+1)t) dt by parts u=t^2 ⇒u′=2t v′=e^(−(n+1)t) ⇒v=((−1)/(n+1))e^(−(n+1)t) A=Σ_(n=0) ^∞ (2/(n+1))∫_0 ^∞ te^(−(n+1)t) dt with ([t^2 e^(−(n+1)t) ]_0 ^∞ =0) by parts A=Σ_(n=0) ^∞ (2/((n+1)^2 ))∫_0 ^∞ e^(−(n+1)t) dt =Σ_(n=0) ^∞ (2/((n+1)^3 ))[e^(−(n+1)t) ]_0 ^∞ =2Σ_(n=0) ^∞ (1/((n+1)^3 )) =2Σ_(n=1) ^∞ (1/n^3 )=2ξ(3)](https://www.tinkutara.com/question/Q53798.png)
$${A}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} }{{e}^{{t}} −\mathrm{1}}{dt}=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} {e}^{−{t}} }{\mathrm{1}−{e}^{−{t}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{2}} {e}^{−{t}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{nt}} {dt} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{2}} {e}^{−\left({n}+\mathrm{1}\right){t}} {dt} \\ $$$${by}\:{parts} \\ $$$${u}={t}^{\mathrm{2}} \Rightarrow{u}'=\mathrm{2}{t} \\ $$$${v}'={e}^{−\left({n}+\mathrm{1}\right){t}} \Rightarrow{v}=\frac{−\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\left({n}+\mathrm{1}\right){t}} \\ $$$${A}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} {te}^{−\left({n}+\mathrm{1}\right){t}} {dt}\:\:{with}\:\:\left(\left[{t}^{\mathrm{2}} {e}^{−\left({n}+\mathrm{1}\right){t}} \right]_{\mathrm{0}} ^{\infty} =\mathrm{0}\right) \\ $$$${by}\:{parts} \\ $$$${A}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} {e}^{−\left({n}+\mathrm{1}\right){t}} {dt} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\left[{e}^{−\left({n}+\mathrm{1}\right){t}} \right]_{\mathrm{0}} ^{\infty} =\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\mathrm{2}\xi\left(\mathrm{3}\right) \\ $$
Commented by maxmathsup by imad last updated on 26/Jan/19
$${thank}\:{you}\:{sir} \\ $$