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Question Number 85009 by mathmax by abdo last updated on 18/Mar/20

calculate ∫_0 ^∞ (x^n /(sh(x)))dx with n integr natural

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{{sh}\left({x}\right)}{dx}\:{with}\:{n}\:{integr}\:{natural} \\ $$

Commented by mathmax by abdo last updated on 18/Mar/20

let A_n =∫_0 ^∞ (x^n /(sh(x)))dx ⇒A_n =2∫_0 ^∞ (x^n /(e^x −e^(−x) ))dx =2 ∫_0 ^∞ ((e^(−x) x^n )/(1−e^(−2x) ))dx =2 ∫_0 ^∞ x^n e^(−x) (Σ_(k=0) ^∞ e^(−2kx) )dx =2 Σ_(k=0) ^∞ ∫_0 ^∞ x^n e^(−(2k+1)x) dx =_((2k+1)x=t) 2Σ_(k=0) ^∞ ∫_0 ^∞ ((t/(2k+1)))^n e^(−t) (dt/((2k+1))) =2 Σ_(k=0) ^∞ (1/((2k+1)^(n+1) ))∫_0 ^∞ t^n e^(−t) dt =2Γ(n+1)Σ_(k=0) ^∞ (1/((2k+1)^(n+1) )) we have ξ(x)=Σ_(k=1) ^∞ (1/k^x ) ⇒ξ(n+1)=Σ_(k=1) ^∞ (1/k^(n+1) ) =Σ_(k=1) ^∞ (1/((2k)^(n+1) ))+Σ_(k=0) ^∞ (1/((2k+1)^(n+1) )) =(1/2^(n+1) )Σ_(k=1) ^∞ (1/k^(n+1) ) +Σ_(k=0) ^∞ (1/((2k+1)^(n+1) )) ⇒Σ_(k=0) ^∞ (1/((2k+1)^(n+1) )) =(1−(1/2^(n+1) ))ξ(n+1) ⇒ A_n =2Γ(n+1)(1−(1/2^(n+1) ))ξ(n+1) =(2−(1/2^n ))(n!)ξ(n+1)

$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{{sh}\left({x}\right)}{dx}\:\Rightarrow{A}_{{n}} =\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{{e}^{{x}} −{e}^{−{x}} }{dx} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{x}} \:{x}^{{n}} }{\mathrm{1}−{e}^{−\mathrm{2}{x}} }{dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:{x}^{{n}} \:{e}^{−{x}} \left(\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{2}{kx}} \right){dx} \\ $$$$=\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{−\left(\mathrm{2}{k}+\mathrm{1}\right){x}} \:{dx}\:=_{\left(\mathrm{2}{k}+\mathrm{1}\right){x}={t}} \:\:\mathrm{2}\sum_{{k}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:\left(\frac{{t}}{\mathrm{2}{k}+\mathrm{1}}\right)^{{n}} \:{e}^{−{t}} \:\frac{{dt}}{\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$=\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} \:\:{t}^{{n}} \:{e}^{−{t}} \:{dt}\:=\mathrm{2}\Gamma\left({n}+\mathrm{1}\right)\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{{n}+\mathrm{1}} } \\ $$$${we}\:{have}\:\xi\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{{x}} }\:\Rightarrow\xi\left({n}+\mathrm{1}\right)=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{{n}+\mathrm{1}} } \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}\right)^{{n}+\mathrm{1}} }+\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{{n}+\mathrm{1}} }\:+\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{{n}+\mathrm{1}} } \\ $$$$\Rightarrow\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\right)\xi\left({n}+\mathrm{1}\right)\:\Rightarrow \\ $$$${A}_{{n}} =\mathrm{2}\Gamma\left({n}+\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\right)\xi\left({n}+\mathrm{1}\right) \\ $$$$=\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\left({n}!\right)\xi\left({n}+\mathrm{1}\right) \\ $$$$ \\ $$

Answered by mind is power last updated on 18/Mar/20

∫_0 ^(+∞) ((2x^n e^x dx)/((e^x −1)(e^x +1))) =∫_0 ^(+∞) ((x^n dx)/(e^x −1))+∫_0 ^(+∞) (x^n /(e^x +1))dx =∫_0 ^(+∞) ((x^n e^(−x) )/(1−e^(−x) ))dx =∫_0 ^(+∞) x^n Σ_(k≥0) e^(−(k+1)x) dx+∫_0 ^(+∞) x^n .Σ_(k≥0) (−1)^k e^(−(1+k)x) dx =Σ_(k≥0) ∫_0 ^(+∞) x^n e^(−(k+1)x) dx =Σ_(k≥0) ∫_0 ^(+∞) (y^n /((k+1)^(n+1) ))e^(−y) dy =Γ(n+1)ζ(n+1) ∫_0 ^(+∞) x^n .Σ_(k≥0) (−1)^k e^(−(1+k)x) dx =Σ_(k≥0) ∫_0 ^(+∞) (y^n /((1+k)^(n+1) ))(−1)^k e^(−y) dy =Γ(n+1)Σ_(k≥0) (−1)^k .(1/((1+k)^(n+1) )) Σ_(k≥0) (((−1)^k )/((1+k)^(n+1) ))=Σ_(k≥0) (1/((1+2k)^(n+1) ))−(1/((2+2k)^(n+1) ))=(1−(1/2^n ))ζ(n+1) =Γ(n+1)(1−(1/2^n ))ζ(n+1) =Γ(n+1){2−(1/2^n )}ζ(n+1)=∫_0 ^(+∞) (x^n /(sh(x)))dx

$$\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{2}{x}^{{n}} {e}^{{x}} {dx}}{\left({e}^{{x}} −\mathrm{1}\right)\left({e}^{{x}} +\mathrm{1}\right)} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{x}^{{n}} {dx}}{{e}^{{x}} −\mathrm{1}}+\int_{\mathrm{0}} ^{+\infty} \frac{{x}^{{n}} }{{e}^{{x}} +\mathrm{1}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{x}^{{n}} {e}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} {x}^{{n}} \underset{{k}\geqslant\mathrm{0}} {\sum}{e}^{−\left({k}+\mathrm{1}\right){x}} {dx}+\int_{\mathrm{0}} ^{+\infty} {x}^{{n}} .\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} {e}^{−\left(\mathrm{1}+{k}\right){x}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{+\infty} {x}^{{n}} {e}^{−\left({k}+\mathrm{1}\right){x}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{+\infty} \frac{{y}^{{n}} }{\left({k}+\mathrm{1}\right)^{{n}+\mathrm{1}} }{e}^{−{y}} {dy} \\ $$$$=\Gamma\left({n}+\mathrm{1}\right)\zeta\left({n}+\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{+\infty} {x}^{{n}} .\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} {e}^{−\left(\mathrm{1}+{k}\right){x}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{+\infty} \frac{{y}^{{n}} }{\left(\mathrm{1}+{k}\right)^{{n}+\mathrm{1}} }\left(−\mathrm{1}\right)^{{k}} {e}^{−{y}} {dy} \\ $$$$=\Gamma\left({n}+\mathrm{1}\right)\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} .\frac{\mathrm{1}}{\left(\mathrm{1}+{k}\right)^{{n}+\mathrm{1}} } \\ $$$$\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{1}+{k}\right)^{{n}+\mathrm{1}} }=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{2}{k}\right)^{{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\left(\mathrm{2}+\mathrm{2}{k}\right)^{{n}+\mathrm{1}} }=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\zeta\left({n}+\mathrm{1}\right) \\ $$$$=\Gamma\left({n}+\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\zeta\left({n}+\mathrm{1}\right) \\ $$$$=\Gamma\left({n}+\mathrm{1}\right)\left\{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right\}\zeta\left({n}+\mathrm{1}\right)=\int_{\mathrm{0}} ^{+\infty} \frac{{x}^{{n}} }{{sh}\left({x}\right)}{dx} \\ $$

Commented by abdomathmax last updated on 18/Mar/20

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

Commented by mind is power last updated on 18/Mar/20

withe pleasur

$${withe}\:{pleasur} \\ $$

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