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Question Number 153721 by mnjuly1970 last updated on 09/Sep/21
prove that : I:= ∫_0 ^( ∞) (( x^( 3) )/(sinh ( x ))) dx = ((π^4 )/8) ■ m.n
$$ \\ $$$$\:\:\:\:\:\mathrm{prove}\:\:\mathrm{that}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{x}^{\:\mathrm{3}} }{{sinh}\:\left(\:{x}\:\right)}\:{dx}\:=\:\frac{\pi\:^{\mathrm{4}} }{\mathrm{8}}\:\:\:\:\:\:\:\:\:\:\blacksquare\:{m}.{n}\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 09/Sep/21
I=∫_0 ^∞ (x^3 /(sh(x)))dx ⇒I=∫_0 ^∞ ((2x^3 )/(e^x −e^(−x) ))dx=2∫_0 ^∞ ((x^3 e^(−x) )/(1−e^(−2x) ))dx =2∫_0 ^∞ x^(3 ) e^(−x) (Σ_(n=0) ^∞ e^(−2nx) )dx =2Σ_(n=0) ^∞ ∫_0 ^∞ x^(3 ) e^(−(2n+1)x) dx =_((2n+1)x=t) 2Σ_(n=0) ^(∞ ) ∫_0 ^∞ (t^3 /((2n+1)^3 ))e^(−t ) (dt/((2n+1))) =2Σ_(n=0) ^(∞ ) (1/((2n+1)^4 ))∫_0 ^∞ t^(3 ) e^(−t) dt =2Γ(4)Σ_(n=0) ^∞ (1/((2n+1)^4 )) Γ(4)=3!=6 Σ_(n=1) ^(∞ ) (1/n^4 )=(π^4 /(90))=(1/2^4 )Σ_(n=1) ^∞ (1/n^4 )+Σ_(n=0) ^∞ (1/((2n+1)^4 )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^4 ))=(1−(1/2^4 ))×(π^4 /(90))=((15)/(16))×(π^4 /(90)) =((3.5π^4 )/(16.3.30))=((5π^4 )/(16.6.5))=(π^4 /(96)) ⇒ I=12×(π^4 /(96))=(π^4 /8)
$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{sh}\left(\mathrm{x}\right)}\mathrm{dx}\:\Rightarrow\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2x}^{\mathrm{3}} }{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }\mathrm{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{x}} }{\mathrm{1}−\mathrm{e}^{−\mathrm{2x}} }\mathrm{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \mathrm{x}^{\mathrm{3}\:} \mathrm{e}^{−\mathrm{x}} \left(\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{2nx}} \right)\mathrm{dx}\:=\mathrm{2}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{3}\:} \mathrm{e}^{−\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x}} \:\mathrm{dx} \\ $$$$=_{\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x}=\mathrm{t}} \:\:\:\:\mathrm{2}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty\:} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{3}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{e}^{−\mathrm{t}\:} \frac{\mathrm{dt}}{\left(\mathrm{2n}+\mathrm{1}\right)} \\ $$$$=\mathrm{2}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty\:} \:\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{4}} }\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{3}\:} \mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:=\mathrm{2}\Gamma\left(\mathrm{4}\right)\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\Gamma\left(\mathrm{4}\right)=\mathrm{3}!=\mathrm{6} \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty\:} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{4}} }=\frac{\pi^{\mathrm{4}} }{\mathrm{90}}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{4}} }+\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{4}} }\:\Rightarrow \\ $$$$\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{4}} }=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\right)×\frac{\pi^{\mathrm{4}} }{\mathrm{90}}=\frac{\mathrm{15}}{\mathrm{16}}×\frac{\pi^{\mathrm{4}} }{\mathrm{90}}\:=\frac{\mathrm{3}.\mathrm{5}\pi^{\mathrm{4}} }{\mathrm{16}.\mathrm{3}.\mathrm{30}}=\frac{\mathrm{5}\pi^{\mathrm{4}} }{\mathrm{16}.\mathrm{6}.\mathrm{5}}=\frac{\pi^{\mathrm{4}} }{\mathrm{96}}\:\Rightarrow \\ $$$$\mathrm{I}=\mathrm{12}×\frac{\pi^{\mathrm{4}} }{\mathrm{96}}=\frac{\pi^{\mathrm{4}} }{\mathrm{8}} \\ $$
Commented by mnjuly1970 last updated on 09/Sep/21
thx sir max
$${thx}\:{sir}\:{max} \\ $$
Answered by Jonathanwaweh last updated on 09/Sep/21
please sir (1/(1−e^(−2x) ))≠Σ_(n=0) ^∞ e^(−2nx) because when x→0 e^(−2x) =1≠0
$${please}\:{sir}\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{2}{x}} }\neq\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−\mathrm{2}{nx}} \:\:{because}\:{when}\:{x}\rightarrow\mathrm{0}\:{e}^{−\mathrm{2}{x}} =\mathrm{1}\neq\mathrm{0} \\ $$
Commented by Mathspace last updated on 09/Sep/21
look sir ∣e^(−2x) ∣<1 and use equality (1/(1−u))=Σu^n ∣u∣<1
$${look}\:{sir}\:\:\mid{e}^{−\mathrm{2}{x}} \mid<\mathrm{1}\:{and}\:{use}\: \\ $$$${equality}\:\frac{\mathrm{1}}{\mathrm{1}−{u}}=\Sigma{u}^{{n}} \:\:\:\:\mid{u}\mid<\mathrm{1} \\ $$

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