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Question Number 127888 by A8;15: last updated on 02/Jan/21

Answered by mindispower last updated on 02/Jan/21

x=sh(t) ⇒dx=ch(t)dt ⇔∫_0 ^∞ sin(ch(t))cos(sh(t))dt..=Ω Λ=∫_0 ^∞ cos(ch(t))sin(sh(t))dt ∫_0 ^∞ sin(e^t )dt=∫_0 ^∞ ((sin(x))/x)dx=(π/2) ∫_0 ^∞ sin(ch(t)+sh(t))dt=∫_0 ^∞ sin(e^t )dt=∫_1 ^∞ ((sin(x))/x)dx Δ=∫_0 ^∞ (sin(ch(t))cos(sh(t))+cos(ch(t))sin(sh(t)))dt Ω+Λ=∫_1 ^∞ ((sin(x))/t)dt Ω−Λ=∫_0 ^∞ sin(e^(−t) )=∫_0 ^1 ((sin(t))/t)dt ⇒2Ω=∫_0 ^1 ((sin(t))/t)dt+∫_1 ^∞ ((sin(t))/t)dt=∫_0 ^∞ ((sin(t))/t)dt=(π/2) ⇔Ω=(π/4)

$${x}={sh}\left({t}\right) \\ $$$$\Rightarrow{dx}={ch}\left({t}\right){dt} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\infty} {sin}\left({ch}\left({t}\right)\right){cos}\left({sh}\left({t}\right)\right){dt}..=\Omega \\ $$$$\Lambda=\int_{\mathrm{0}} ^{\infty} {cos}\left({ch}\left({t}\right)\right){sin}\left({sh}\left({t}\right)\right){dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} {sin}\left({e}^{{t}} \right){dt}=\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{dx}=\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} {sin}\left({ch}\left({t}\right)+{sh}\left({t}\right)\right){dt}=\int_{\mathrm{0}} ^{\infty} {sin}\left({e}^{{t}} \right){dt}=\int_{\mathrm{1}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{dx} \\ $$$$\Delta=\int_{\mathrm{0}} ^{\infty} \left({sin}\left({ch}\left({t}\right)\right){cos}\left({sh}\left({t}\right)\right)+{cos}\left({ch}\left({t}\right)\right){sin}\left({sh}\left({t}\right)\right)\right){dt} \\ $$$$\Omega+\Lambda=\int_{\mathrm{1}} ^{\infty} \frac{{sin}\left({x}\right)}{{t}}{dt} \\ $$$$\Omega−\Lambda=\int_{\mathrm{0}} ^{\infty} {sin}\left({e}^{−{t}} \right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({t}\right)}{{t}}{dt} \\ $$$$\Rightarrow\mathrm{2}\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({t}\right)}{{t}}{dt}+\int_{\mathrm{1}} ^{\infty} \frac{{sin}\left({t}\right)}{{t}}{dt}=\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({t}\right)}{{t}}{dt}=\frac{\pi}{\mathrm{2}} \\ $$$$\Leftrightarrow\Omega=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by A8;15: last updated on 03/Jan/21

thanks sir. Happy New Year

$$\mathrm{thanks}\:\mathrm{sir}.\:\boldsymbol{\mathrm{Happy}}\:\boldsymbol{\mathrm{New}}\:\boldsymbol{\mathrm{Year}} \\ $$

Commented by mindispower last updated on 07/Jan/21

withe pleasur

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

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