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Ω= ∫_0 ^( ∞) (( x − sin (x ))/x^( 3) )dx −−− solution−−− Ω=^(I.B.P) [ ((−1)/(2 x^( 2) )) (x−sin(x))]_0 ^∞ +(1/2) ∫_0 ^( ∞) ((1−cos (x))/x^( 2) )dx = (1/2) ∫_0 ^( ∞) (( 2sin^( 2) ((x/2)))/x^( 2) )dx=∫_0 ^( ∞) ((sin^( 2) ((x/2)))/x^( 2) )dx =^((x/2) = α) (1/2)∫_0 ^( ∞) ((sin^( 2) ( α))/α^( 2) ) dα = (1/2) [((−1)/α) sin^( 2) (α)]_0 ^∞ +(1/2)∫_0 ^( ∞) ((sin(2α))/α)dα =^(2α=ϕ) (1/2) ∫_0 ^( ∞) (( sin(ϕ ))/ϕ) dϕ =(π/4) −− Ω= (π/4) −−−

$$ \\ $$$$\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{x}\:−\:{sin}\:\left({x}\:\right)}{{x}^{\:\mathrm{3}} }{dx} \\ $$$$−−−\:{solution}−−− \\ $$$$\:\:\:\:\:\Omega\overset{\mathscr{I}.\mathscr{B}.\mathscr{P}} {=}\:\left[\:\frac{−\mathrm{1}}{\mathrm{2}\:{x}^{\:\mathrm{2}} }\:\left({x}−{sin}\left({x}\right)\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−{cos}\:\left({x}\right)}{{x}^{\:\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:\mathrm{2}{sin}^{\:\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{{x}^{\:\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\:\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{{x}^{\:\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\overset{\frac{{x}}{\mathrm{2}}\:=\:\alpha} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\:\mathrm{2}} \left(\:\alpha\right)}{\alpha^{\:\mathrm{2}} }\:{d}\alpha\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\frac{−\mathrm{1}}{\alpha}\:{sin}^{\:\mathrm{2}} \left(\alpha\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\mathrm{2}\alpha\right)}{\alpha}{d}\alpha \\ $$$$\:\:\:\:\:\:\:\:\overset{\mathrm{2}\alpha=\varphi} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{sin}\left(\varphi\:\right)}{\varphi}\:{d}\varphi\:=\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−\:\:\:\:\:\Omega=\:\frac{\pi}{\mathrm{4}}\:\:−−− \\ $$$$ \\ $$$$ \\ $$

Question Number 163002 Answers: 0 Comments: 0

prove that i:Σ_(n=0) ^∞ (((−1 )^( n) )/((n +(1/2))cosh(n+(1/2))π)) =(π/4) ii: ∫_0 ^( 1) (( sin( π x ))/(x^( x) ( 1−x )^( 1−x) )) (dx/(1+x)) =(π/4) −−−

$$ \\ $$$$\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\:{i}:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\:\right)^{\:{n}} }{\left({n}\:+\frac{\mathrm{1}}{\mathrm{2}}\right){cosh}\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\pi}\:=\frac{\pi}{\mathrm{4}} \\ $$$$\:\:{ii}:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{sin}\left(\:\pi\:{x}\:\right)}{{x}^{\:{x}} \left(\:\mathrm{1}−{x}\:\right)^{\:\mathrm{1}−{x}} }\:\frac{{dx}}{\mathrm{1}+{x}}\:=\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:−−− \\ $$

Question Number 162939 Answers: 0 Comments: 0

lim_( x→ 3) ( a ⌊x ⌋ + ⌊ −x⌋).tan(((πx)/2) )=−∞ a ∈ ?

$$ \\ $$$$\:\:{lim}_{\:{x}\rightarrow\:\mathrm{3}} \:\left(\:{a}\:\lfloor{x}\:\rfloor\:+\:\lfloor\:−{x}\rfloor\right).{tan}\left(\frac{\pi{x}}{\mathrm{2}}\:\right)=−\infty \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{a}\:\in\:? \\ $$$$ \\ $$

Question Number 162924 Answers: 2 Comments: 0

𝛗 =∫_0 ^( ∞) (( e^( −x^( 2) ) .ln( x ))/( (√x))) dx=λ Γ((1/4)) λ=? ■

$$\: \\ $$$$\:\boldsymbol{\phi}\:=\int_{\mathrm{0}} ^{\:\infty} \frac{\:{e}^{\:−{x}^{\:\mathrm{2}} } .\mathrm{ln}\left(\:{x}\:\right)}{\:\sqrt{{x}}}\:{dx}=\lambda\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\lambda=?\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare \\ $$$$ \\ $$

Question Number 162804 Answers: 2 Comments: 0

Ω = ∫ sin^( 2) (x).cos^( 4) (x ) dx

$$ \\ $$$$ \\ $$$$\:\:\:\Omega\:=\:\int\:{sin}^{\:\mathrm{2}} \left({x}\right).{cos}^{\:\mathrm{4}} \left({x}\:\right)\:{dx} \\ $$$$ \\ $$

Question Number 162701 Answers: 1 Comments: 0