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Question Number 133107 by mnjuly1970 last updated on 18/Feb/21

        ... nice   calculus...     find::   𝛗=^(???) ∫_0 ^( 1) (sin(x)+sin((1/x)))(dx/x)

$$\:\:\:\:\:\:\:\:...\:{nice}\:\:\:{calculus}... \\ $$$$\:\:\:{find}::\:\:\:\boldsymbol{\phi}\overset{???} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({sin}\left({x}\right)+{sin}\left(\frac{\mathrm{1}}{{x}}\right)\right)\frac{{dx}}{{x}} \\ $$

Answered by Dwaipayan Shikari last updated on 18/Feb/21

Φ=∫_0 ^1 (sin(x)+sin((1/x)))(dx/x)                  (1/x)=u⇒−(1/x^2 )=(du/dx)  =∫_1 ^∞ ((sin((1/u)))/u)+((sinu)/u)du=∫_0 ^∞ ((sin((1/u)))/u)+((sinu)/u)du−Φ  2Φ=∫_0 ^∞ ((sinu)/u)+(1/u).sin((1/u))du  Φ=(π/4)+(1/2)∫_0 ^∞ ((sin(t))/t) du               (1/u)=t⇒−(1/u^2 )=(dt/du)  =(π/4)+(π/4)=(π/2)

$$\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} \left({sin}\left({x}\right)+{sin}\left(\frac{\mathrm{1}}{{x}}\right)\right)\frac{{dx}}{{x}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{x}}={u}\Rightarrow−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\frac{{du}}{{dx}} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{{sin}\left(\frac{\mathrm{1}}{{u}}\right)}{{u}}+\frac{{sinu}}{{u}}{du}=\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\frac{\mathrm{1}}{{u}}\right)}{{u}}+\frac{{sinu}}{{u}}{du}−\Phi \\ $$$$\mathrm{2}\Phi=\int_{\mathrm{0}} ^{\infty} \frac{{sinu}}{{u}}+\frac{\mathrm{1}}{{u}}.{sin}\left(\frac{\mathrm{1}}{{u}}\right){du} \\ $$$$\Phi=\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({t}\right)}{{t}}\:{du}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{u}}={t}\Rightarrow−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }=\frac{{dt}}{{du}} \\ $$$$=\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 19/Feb/21

grateful mr payan...

$${grateful}\:{mr}\:{payan}... \\ $$

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