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Question Number 130949 by mnjuly1970 last updated on 31/Jan/21
...nice calculus... calculate: I=∫_0 ^( ∞) ((Si(x))/x^(3/2) ) dx=^(???) (√(8π)) Si(x)=∫_0 ^( x) ((sin(t))/t)dt
$$\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:\:\:{calculate}:\: \\ $$$$\:\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{S}{i}\left({x}\right)}{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:{dx}\overset{???} {=}\sqrt{\mathrm{8}\pi}\: \\ $$$$\:\:\:\:\:\:\:\mathrm{S}{i}\left({x}\right)=\int_{\mathrm{0}} ^{\:{x}} \frac{{sin}\left({t}\right)}{{t}}{dt} \\ $$
Answered by mindispower last updated on 31/Jan/21
∫((si(x))/x^(3/2) )dx=−((2si(x))/( (√x)))+2∫((sin(x))/( (√x).x))dx put x=u^2 ,in 2nd 2∫((sin(x))/(x(√x)))dx=4∫((sin(u^2 ))/u^2 )du by part=−((4sin(u^2 ))/u)+8∫cos(u^2 )du...E ∫_0 ^t cos((π/2)x^2 )dx=C(t) fersnel integral u=(√((π/2) ))y ∫(√(π/2))ycos(((πy^2 )/2))dy=(√(π/2))C(y) we get E=−((4sin(u^2 ))/u)+4(√(2π))C((√(2/π))u) ∫((si(x))/x^(3/2) )dx=((−2si(x))/( (√x)))−((4sin(x))/( (√x)))+4(√(2π)).C((√(2/π)).(√x)) x→∞ lim_(x→∞) C(x)=(1/2) ⇒∫_0 ^∞ ((si(x))/(x(√x)))dx=lim_(x→∞) (−((2si(x))/( (√x)))−((4sin(x))/( (√x)))+4(√(2π))Ci((√((2x)/π)))) =2(√(2π))=(√(8π))
$$\int\frac{{si}\left({x}\right)}{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}=−\frac{\mathrm{2}{si}\left({x}\right)}{\:\sqrt{{x}}}+\mathrm{2}\int\frac{{sin}\left({x}\right)}{\:\sqrt{{x}}.{x}}{dx} \\ $$$${put}\:{x}={u}^{\mathrm{2}} ,{in}\:\mathrm{2}{nd} \\ $$$$\mathrm{2}\int\frac{{sin}\left({x}\right)}{{x}\sqrt{{x}}}{dx}=\mathrm{4}\int\frac{{sin}\left({u}^{\mathrm{2}} \right)}{{u}^{\mathrm{2}} }{du} \\ $$$${by}\:{part}=−\frac{\mathrm{4}{sin}\left({u}^{\mathrm{2}} \right)}{{u}}+\mathrm{8}\int{cos}\left({u}^{\mathrm{2}} \right){du}…{E} \\ $$$$\int_{\mathrm{0}} ^{{t}} {cos}\left(\frac{\pi}{\mathrm{2}}{x}^{\mathrm{2}} \right){dx}={C}\left({t}\right)\:{fersnel}\:{integral} \\ $$$${u}=\sqrt{\frac{\pi}{\mathrm{2}}\:}{y} \\ $$$$\int\sqrt{\frac{\pi}{\mathrm{2}}}{ycos}\left(\frac{\pi{y}^{\mathrm{2}} }{\mathrm{2}}\right){dy}=\sqrt{\frac{\pi}{\mathrm{2}}}{C}\left({y}\right) \\ $$$${we}\:{get} \\ $$$${E}=−\frac{\mathrm{4}{sin}\left({u}^{\mathrm{2}} \right)}{{u}}+\mathrm{4}\sqrt{\mathrm{2}\pi}{C}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{u}\right) \\ $$$$\int\frac{{si}\left({x}\right)}{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}=\frac{−\mathrm{2}{si}\left({x}\right)}{\:\sqrt{{x}}}−\frac{\mathrm{4}{sin}\left({x}\right)}{\:\sqrt{{x}}}+\mathrm{4}\sqrt{\mathrm{2}\pi}.{C}\left(\sqrt{\frac{\mathrm{2}}{\pi}}.\sqrt{{x}}\right) \\ $$$${x}\rightarrow\infty \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{C}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{si}\left({x}\right)}{{x}\sqrt{{x}}}{dx}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(−\frac{\mathrm{2}{si}\left({x}\right)}{\:\sqrt{{x}}}−\frac{\mathrm{4}{sin}\left({x}\right)}{\:\sqrt{{x}}}+\mathrm{4}\sqrt{\mathrm{2}\pi}{Ci}\left(\sqrt{\frac{\mathrm{2}{x}}{\pi}}\right)\right) \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}\pi}=\sqrt{\mathrm{8}\pi} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 31/Jan/21
very nice as always sir mindispower... Allah keep you..
$${very}\:{nice}\:{as}\:{always} \\ $$$${sir}\:\:{mindispower}… \\ $$$$\:{Allah}\:{keep}\:{you}.. \\ $$
Commented by mindispower last updated on 31/Jan/21
inshallah alaways withe pleasur
$${inshallah}\:{alaways}\:{withe}\:{pleasur} \\ $$

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