Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 117948 by mnjuly1970 last updated on 14/Oct/20

... nice integral... please evaluate :: I =∫_0 ^( 1) (sin(x)+sin((1/x)))(dx/x) =?? m.n.1970

$$\:\:\:\:\:\:\:\:...\:\:{nice}\:\:{integral}...\: \\ $$$$\:\:\:{please}\:{evaluate}\::: \\ $$$$ \\ $$$$\:\:\:\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({sin}\left({x}\right)+{sin}\left(\frac{\mathrm{1}}{{x}}\right)\right)\frac{{dx}}{{x}}\:=?? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$$$\:\: \\ $$

Answered by Lordose last updated on 14/Oct/20

I=∫_( 0) ^( 1) ((sinx)/x)dx + ∫_( 0) ^( 1) ((sin((1/x)))/x)dx I= ∣Si(x)∣_0 ^1 −∣Si((1/x))∣_0 ^1 I= Si(1)−Si(0) − Si(1) + Si(∞) I= (π/2)

$$ \\ $$$$\mathrm{I}=\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}\:+\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{I}=\:\mid\mathrm{Si}\left(\mathrm{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \:−\mid\mathrm{Si}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\mathrm{I}=\:\mathrm{Si}\left(\mathrm{1}\right)−\mathrm{Si}\left(\mathrm{0}\right)\:−\:\mathrm{Si}\left(\mathrm{1}\right)\:+\:\mathrm{Si}\left(\infty\right) \\ $$$$\mathrm{I}=\:\frac{\pi}{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 14/Oct/20

grateful..

$${grateful}.. \\ $$

Answered by AbduraufKodiriy last updated on 14/Oct/20

Commented by mnjuly1970 last updated on 14/Oct/20

thank you ..

$${thank}\:{you}\:.. \\ $$

Answered by 1549442205PVT last updated on 14/Oct/20

I=∫_0 ^( 1) ((sinx)/x)dx+∫_0 ^( 1) ((sin(1/x))/x)dx =A+B A=∫_0 ^( 1) ((sint)/t)dt Put t=(1/x)⇒dt=−(1/x^2 )dx=−t^2 dx B=−∫_∞ ^( 1) ((tsint)/t^2 )dt=∫_1 ^∞ ((sint)/t)dt ⇒I=A+B=∫_0 ^( 1) ((sint)/t)dt+∫_1 ^∞ ((sint)/t)dt =∫_0 ^∞ ((sint)/t)=(π/2) (this is proved ago)

$$\mathrm{I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}+\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{sin}\frac{\mathrm{1}}{\mathrm{x}}}{\mathrm{x}}\mathrm{dx}\:=\mathrm{A}+\mathrm{B} \\ $$$$\mathrm{A}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt} \\ $$$$\mathrm{Put}\:\mathrm{t}=\frac{\mathrm{1}}{\mathrm{x}}\Rightarrow\mathrm{dt}=−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=−\mathrm{t}^{\mathrm{2}} \mathrm{dx} \\ $$$$\mathrm{B}=−\int_{\infty} ^{\:\mathrm{1}} \frac{\mathrm{tsint}}{\mathrm{t}^{\mathrm{2}} }\mathrm{dt}=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt} \\ $$$$\Rightarrow\mathrm{I}=\mathrm{A}+\mathrm{B}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt}+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sint}}{\mathrm{t}}=\frac{\pi}{\mathrm{2}}\:\left(\mathrm{this}\:\mathrm{is}\:\mathrm{proved}\:\mathrm{ago}\right) \\ $$

Commented by mnjuly1970 last updated on 14/Oct/20

mercey..

$${mercey}.. \\ $$

Answered by mindispower last updated on 14/Oct/20

I=∫_0 ^1 ((sin(x))/x)dx+∫_0 ^1 ((sin((1/x)))/x)dx let (1/x)=t in 2nd ⇒ ∫_0 ^1 ((sin((1/x)))/x)dx=∫_1 ^∞ ((sin(t))/t)dt I=∫_0 ^1 ((sin(t))/t)dt+∫_1 ^∞ ((sin(t))/t)dt=∫_0 ^∞ ((sin(t))/t)dt=(π/2)

$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({x}\right)}{{x}}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}}{dx} \\ $$$${let}\:\frac{\mathrm{1}}{{x}}={t}\:{in}\:\mathrm{2}{nd}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}}{dx}=\int_{\mathrm{1}} ^{\infty} \frac{{sin}\left({t}\right)}{{t}}{dt} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({t}\right)}{{t}}{dt}+\int_{\mathrm{1}} ^{\infty} \frac{{sin}\left({t}\right)}{{t}}{dt}=\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({t}\right)}{{t}}{dt}=\frac{\pi}{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 14/Oct/20

very nice as always mr power..

$${very}\:{nice}\:{as}\:{always}\:{mr} \\ $$$${power}.. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com