Question-154143

Question Number 154143 by mnjuly1970 last updated on 14/Sep/21

Answered by phanphuoc last updated on 14/Sep/21

u=−lnx→e^(−u) =x→−e^(−u) du=dx Ω=∫_∞ ^0 ((e^(−x) sinx)/(1+e^(−2x) ))(−e^(−x) dx)=∫_0 ^∞ ((e^(−2x) sinxdx)/(1+e^(−2x) )) =∫_0 ^∞ e^(−2x) sinxΣ(−1)^n e^(−2nx) dx= Σ(−1)^n ∫_0 ^∞ e^(−2x(n+1)) sinxdx_=

$${u}=−{lnx}\rightarrow{e}^{−{u}} ={x}\rightarrow−{e}^{−{u}} {du}={dx} \\ $$$$\Omega=\int_{\infty} ^{\mathrm{0}} \frac{{e}^{−{x}} {sinx}}{\mathrm{1}+{e}^{−\mathrm{2}{x}} }\left(−{e}^{−{x}} {dx}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\mathrm{2}{x}} {sinxdx}}{\mathrm{1}+{e}^{−\mathrm{2}{x}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} {sinx}\Sigma\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{nx}}} \boldsymbol{{dx}}= \\ $$$$\Sigma\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}\left({n}+\mathrm{1}\right)} {sinxdx}_{=} \\ $$

Answered by puissant last updated on 14/Sep/21

Ω=∫_0 ^1 ((xsin(lnx))/(1+x^2 ))dx u=−lnx → x=e^(−u) → dx=−e^(−u) du Ω=−∫_∞ ^0 ((e^(−u) sinu)/(1+e^(−2u) ))(−e^(−u) du)=−∫_0 ^∞ ((e^(−2u) sinu)/(1+e^(−2u) ))du =−∫_0 ^∞ ((sinu)/(1+e^(2u) ))du = ∫_0 ^∞ Σ_(n=1) ^∞ (−1)^n e^(−2nu) sinudu =Σ_(n=1) ^∞ (−1)^n ∫_0 ^∞ e^(−2nu) sinu du = Σ_(n=1) ^∞ (−1)^n ×(1/(4n^2 +1)) =Σ_(n=1) ^∞ (((−1)^n )/(4n^2 +1)) = (π/4)csch((π/2))−(1/2).. ∴∵ Ω = ((πcsch((π/2))−2)/4)... ................Le puissant..............

$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xsin}\left({lnx}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${u}=−{lnx}\:\rightarrow\:{x}={e}^{−{u}} \:\rightarrow\:{dx}=−{e}^{−{u}} {du} \\ $$$$\Omega=−\int_{\infty} ^{\mathrm{0}} \frac{{e}^{−{u}} {sinu}}{\mathrm{1}+{e}^{−\mathrm{2}{u}} }\left(−{e}^{−{u}} {du}\right)=−\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\mathrm{2}{u}} {sinu}}{\mathrm{1}+{e}^{−\mathrm{2}{u}} }{du} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \frac{{sinu}}{\mathrm{1}+{e}^{\mathrm{2}{u}} }{du}\:=\:\int_{\mathrm{0}} ^{\infty} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {e}^{−\mathrm{2}{nu}} {sinudu} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{nu}} {sinu}\:{du}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} ×\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}}\:=\:\frac{\pi}{\mathrm{4}}{csch}\left(\frac{\pi}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\therefore\because\:\:\:\:\:\:\Omega\:=\:\frac{\pi{csch}\left(\frac{\pi}{\mathrm{2}}\right)−\mathrm{2}}{\mathrm{4}}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…………….\mathscr{L}{e}\:{puissant}………….. \\ $$

Commented by mnjuly1970 last updated on 15/Sep/21

$${grate}.. \\ $$

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