Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 133518 by Abdoulaye last updated on 22/Feb/21

Answered by mnjuly1970 last updated on 22/Feb/21

ans: (π^3 /(16))

$${ans}:\:\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\ $$

Answered by Dwaipayan Shikari last updated on 22/Feb/21

∫_(−∞) ^0 ((t^2 e^t )/(e^(2t) +1))dt log(x)=t⇒1=x(dt/dx) =∫_0 ^∞ ((u^2 e^u )/(1+e^(2u) )) dt t=−u =Σ_(n=1) ^∞ (−1)^(n+1) ∫_0 ^∞ e^(−2nu+u) u^2 du =Σ_(n=1) ^∞ (((−1)^(n+1) )/((2n−1)^3 ))Γ(3)=2Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 ))=(π^3 /(16))

$$\int_{−\infty} ^{\mathrm{0}} \frac{{t}^{\mathrm{2}} {e}^{{t}} }{{e}^{\mathrm{2}{t}} +\mathrm{1}}{dt}\:\:\:\:\:\:\:\:\:\:{log}\left({x}\right)={t}\Rightarrow\mathrm{1}={x}\frac{{dt}}{{dx}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\mathrm{2}} {e}^{{u}} \:\:}{\mathrm{1}+{e}^{\mathrm{2}{u}} }\:{dt}\:\:\:\:\:\:{t}=−{u} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{nu}+{u}} {u}^{\mathrm{2}} {du} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} }\Gamma\left(\mathrm{3}\right)=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\ $$

Commented by Dwaipayan Shikari last updated on 22/Feb/21

(π/2)tan((π/2)x)=(1/((1−x)))−(1/((1+x)))+(1/((3−x)))−(1/((3+x)))+... Differentiating two times respect to x 2.(π^3 /8)sec^2 ((π/2)x)tan((π/2)x)=(1/((1−x)^3 ))−(1/((1+x)^3 ))+(1/((3−x)^3 ))−(1/((3+x)^3 ))+.. x=(1/2) (π^3 /(32))=1−(1/3^3 )+(1/5^3 )−(1/7^3 )+...

$$\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)}+\frac{\mathrm{1}}{\left(\mathrm{3}−{x}\right)}−\frac{\mathrm{1}}{\left(\mathrm{3}+{x}\right)}+... \\ $$$${Differentiating}\:{two}\:{times}\:{respect}\:{to}\:{x} \\ $$$$\mathrm{2}.\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}{x}\right){tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{3}−{x}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{3}+{x}\right)^{\mathrm{3}} }+.. \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\pi^{\mathrm{3}} }{\mathrm{32}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{3}} }+... \\ $$

Commented by Abdoulaye last updated on 22/Feb/21

what it′s Γ?

$${what}\:{it}'{s}\:\Gamma? \\ $$

Commented by Dwaipayan Shikari last updated on 22/Feb/21

Gamma function Γ(a)=∫_0 ^∞ t^(a−1) e^(−t) dt

$${Gamma}\:{function} \\ $$$$\Gamma\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{a}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$

Commented by Abdoulaye last updated on 22/Feb/21

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Commented by Abdoulaye last updated on 23/Feb/21

how did you find (Π/2)tan((Π/2)x)=(1/((1−x)))−(1/((1+x)))+(1/((3−x)))−(1/((3+x)))+... ?

$${how}\:{did}\:{you}\:{find}\:\frac{\Pi}{\mathrm{2}}{tan}\left(\frac{\Pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)}+\frac{\mathrm{1}}{\left(\mathrm{3}−{x}\right)}−\frac{\mathrm{1}}{\left(\mathrm{3}+{x}\right)}+...\:\:? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com