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Question Number 213138 by Spillover last updated on 31/Oct/24
Commented by Frix last updated on 31/Oct/24
We must first solve ∫sin t^3 dt which might be possible using sin t^3 =((e^(it^3 ) −e^(−it^3 ) )/(2i))=((1−e^(2t^3 ) )/(2e^t^3 ))i But how? Then ∫sin t^2 dt (Fresnel−Integral) I don′t think many here among us can solve this. Approximating I get ∫_0 ^1 ∫_0 ^(1/2) ∫_0 ^(1/3) sin (x+y^2 +z^3 ) dz dy dx≈.088997
$$\mathrm{We}\:\mathrm{must}\:\mathrm{first}\:\mathrm{solve}\:\int\mathrm{sin}\:{t}^{\mathrm{3}} \:{dt}\:\mathrm{which}\:\mathrm{might} \\ $$$$\mathrm{be}\:\mathrm{possible}\:\mathrm{using} \\ $$$$\mathrm{sin}\:{t}^{\mathrm{3}} \:=\frac{\mathrm{e}^{\mathrm{i}{t}^{\mathrm{3}} } −\mathrm{e}^{−\mathrm{i}{t}^{\mathrm{3}} } }{\mathrm{2i}}=\frac{\mathrm{1}−\mathrm{e}^{\mathrm{2}{t}^{\mathrm{3}} } }{\mathrm{2e}^{{t}^{\mathrm{3}} } }\mathrm{i} \\ $$$$\mathrm{But}\:\mathrm{how}? \\ $$$$\mathrm{Then}\:\int\mathrm{sin}\:{t}^{\mathrm{2}} \:{dt}\:\left(\mathrm{Fresnel}−\mathrm{Integral}\right) \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{many}\:\mathrm{here}\:\mathrm{among}\:\mathrm{us}\:\mathrm{can}\:\mathrm{solve} \\ $$$$\mathrm{this}. \\ $$$$\mathrm{Approximating}\:\mathrm{I}\:\mathrm{get} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\:\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{3}}} {\int}}\mathrm{sin}\:\left({x}+{y}^{\mathrm{2}} +{z}^{\mathrm{3}} \right)\:{dz}\:{dy}\:{dx}\approx.\mathrm{088997} \\ $$
Answered by MrGaster last updated on 31/Oct/24
=∫_0 ^1 ∫_0 ^(1/2) [−((cos(x+y^2 +z^3 ))/(3z^2 ))]_0 ^(1/3) dydx =∫_0 ^1 ∫_0 ^(1/2) (−((cos(x+y^2 +(1/(27)))/(3∙(1/9)))+((cos(x+y^2 ))/(3∙0^2 )))dydx =∫_0 ^1 ∫_0 ^(1/2) (−3cos(x+y^2 +(1/(27)))+cos2(x+y^2 ))dydx =∫_0 ^1 [sin(x+y^2 )−3sin(x+y^2 +(1/(27)))]_0 ^(1/2) dx =∫_0 ^1 (sin(x+(1/4))−3sin(x+(1/4)+(1/(27)))−sin(x)+3sin(x+(1/(27))))dx =[−cos(x+(1/4))−3sin(x+(1/4)+(1/(27)))+cos(x)−3cos(x+(1/(27)))]_0 ^1 =−cos(1+(1/4))+3cos(1+(1/4)+(1/(27)))+cos(1)−3cos(1+(1/(27)))+cos((1/4))−3cos((1/4)+(1/(27)))−cos(0)+cos((1/(27)))
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left[−\frac{\mathrm{cos}\left({x}+{y}^{\mathrm{2}} +{z}^{\mathrm{3}} \right)}{\mathrm{3}{z}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {dydx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(−\frac{\mathrm{cos}\left({x}+{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}\right.}{\mathrm{3}\centerdot\frac{\mathrm{1}}{\mathrm{9}}}+\frac{\mathrm{cos}\left({x}+{y}^{\mathrm{2}} \right)}{\mathrm{3}\centerdot\mathrm{0}^{\mathrm{2}} }\right){dydx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(−\mathrm{3cos}\left({x}+{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}\right)+\mathrm{cos2}\left({x}+{y}^{\mathrm{2}} \right)\right){dydx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\mathrm{sin}\left({x}+{y}^{\mathrm{2}} \right)−\mathrm{3sin}\left({x}+{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}\right)\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{sin}\left({x}+\frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{3sin}\left({x}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}\right)−\mathrm{sin}\left({x}\right)+\mathrm{3sin}\left({x}+\frac{\mathrm{1}}{\mathrm{27}}\right)\right){dx} \\ $$$$=\left[−\mathrm{cos}\left({x}+\frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{3sin}\left({x}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}\right)+\mathrm{cos}\left({x}\right)−\mathrm{3cos}\left({x}+\frac{\mathrm{1}}{\mathrm{27}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\mathrm{cos}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{3cos}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}\right)+\mathrm{cos}\left(\mathrm{1}\right)−\mathrm{3cos}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{27}}\right)+\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{3cos}\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}\right)−\mathrm{cos}\left(\mathrm{0}\right)+\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{27}}\right) \\ $$
Commented by Frix last updated on 31/Oct/24
No.
$$\mathrm{No}. \\ $$
Answered by Berbere last updated on 31/Oct/24
Im ∫_0 ^1 ∫_0 ^(1/2) ∫_0 ^(1/3) e^(i(x+y^2 +z^3 )) dxdydz =Im∫_0 ^1 e^(ix) ∫_0 ^(1/2) e^(iy^2 ) dy∫_0 ^(1/3) e^(iz^3 ) dz =Im[−ie^i +i].A.B −y=x^n ⇒x=y^(1/n) e^(i(π/n)) =(ye^(iπ) )^(1/n) ∫e^x^n dx=∫e^(−y) .(1/n)e^(i(π/n)) y^((1/n)−1) dy=−(1/n)e^(i(π/n)) Γ((1/n),y) Γ(a,z)=∫_z ^∞ t^(a−1) e^(−t) dt [incomplete Gamma function] −(1/n)e^(i(π/n)) Γ((1/n),−x^n )+c ⇒A=∫_0 ^((1/2) ) e^(iy^2 ) dy=(1/2)∫_0 e^(−z) i^(1/2) z^((1/2)−1) dz =−(1/2)i^(1/2) Γ((1/2),z) A=−(1/2)i^(1/2) Γ((1/2),−iy^2 )]_0 ^(1/2) =−(1/2)e^((iπ)/4) Γ((1/2),−(i/4)]+(1/2)e^(i(π/4)) Γ((1/2),0] B=−(1/3)i^(1/3) Γ((1/3),−(i/(27)))+(i^(1/3) /3)Γ((1/3),0) ∫_0 ^1 ∫_0 ^(1/2) ∫_0 ^(1/3) sin(x+y^2 +z^3 )dxdydz =TM(i(1−e^i )(−(1/2)e^(i(π/4)) (Γ((1/2),−(i/4))+Γ((1/2),0)).(−(1/3)e^(i(π/6)) Γ((1/3),−(i/(27)),)+(e^(i(π/6)) /3)Γ((1/3),0)) IM imaginary Part
$${Im}\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {e}^{{i}\left({x}+{y}^{\mathrm{2}} +{z}^{\mathrm{3}} \right)} {dxdydz} \\ $$$$={Im}\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{ix}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{iy}^{\mathrm{2}} } {dy}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {e}^{{iz}^{\mathrm{3}} } {dz} \\ $$$$={Im}\left[−{ie}^{{i}} +{i}\right].{A}.{B} \\ $$$$−{y}={x}^{{n}} \Rightarrow{x}={y}^{\frac{\mathrm{1}}{{n}}} {e}^{{i}\frac{\pi}{{n}}} =\left({ye}^{{i}\pi} \right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\int{e}^{{x}^{{n}} } {dx}=\int{e}^{−{y}} .\frac{\mathrm{1}}{{n}}{e}^{{i}\frac{\pi}{{n}}} {y}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} {dy}=−\frac{\mathrm{1}}{{n}}{e}^{{i}\frac{\pi}{{n}}} \Gamma\left(\frac{\mathrm{1}}{{n}},{y}\right) \\ $$$$\Gamma\left({a},{z}\right)=\int_{{z}} ^{\infty} {t}^{{a}−\mathrm{1}} {e}^{−{t}} {dt}\:\left[{incomplete}\:{Gamma}\:{function}\right] \\ $$$$−\frac{\mathrm{1}}{{n}}{e}^{{i}\frac{\pi}{{n}}} \Gamma\left(\frac{\mathrm{1}}{{n}},−{x}^{{n}} \right)+{c} \\ $$$$\Rightarrow{A}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}\:} \:{e}^{{iy}^{\mathrm{2}} } {dy}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} {e}^{−{z}} {i}^{\frac{\mathrm{1}}{\mathrm{2}}} {z}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dz} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{i}^{\frac{\mathrm{1}}{\mathrm{2}}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},{z}\right) \\ $$$$\left.{A}=−\frac{\mathrm{1}}{\mathrm{2}}{i}^{\frac{\mathrm{1}}{\mathrm{2}}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},−{iy}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} =−\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},−\frac{{i}}{\mathrm{4}}\right]+\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},\mathrm{0}\right] \\ $$$${B}=−\frac{\mathrm{1}}{\mathrm{3}}{i}^{\frac{\mathrm{1}}{\mathrm{3}}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{3}},−\frac{{i}}{\mathrm{27}}\right)+\frac{{i}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{3}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}},\mathrm{0}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {sin}\left({x}+{y}^{\mathrm{2}} +{z}^{\mathrm{3}} \right){dxdydz} \\ $$$$=\mathcal{TM}\left({i}\left(\mathrm{1}−{e}^{{i}} \right)\left(−\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},−\frac{{i}}{\mathrm{4}}\right)+\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},\mathrm{0}\right)\right).\left(−\frac{\mathrm{1}}{\mathrm{3}}{e}^{{i}\frac{\pi}{\mathrm{6}}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{3}},−\frac{{i}}{\mathrm{27}},\right)+\frac{{e}^{{i}\frac{\pi}{\mathrm{6}}} }{\mathrm{3}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}},\mathrm{0}\right)\right)\right.\right. \\ $$$$\mathcal{IM}\:{imaginary}\:{Part} \\ $$
Commented by Frix last updated on 31/Oct/24
I was hoping you show us the solution, thank you!
$$\mathrm{I}\:\mathrm{was}\:\mathrm{hoping}\:\mathrm{you}\:\mathrm{show}\:\mathrm{us}\:\mathrm{the}\:\mathrm{solution}, \\ $$$$\mathrm{thank}\:\mathrm{you}! \\ $$

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