Question

$\displaystyle \sf \red{ \int^{ \infty}_{0} {e}^{ - {x}^{3} } sin( {x}^{3} ) \: dx }$​

Answer 1

Answer:

here is your answer.

I am trying to show

∫∞0sin3(x)x3dx=3π8.

I believe the contour I should use is a semicircle in the upper half plane with a slight bump at the origin, so I miss the singularity.

Lastly I have the hint to consider

∫∞0e3iz−3eiz+2z3dz

around the contour I mentioned.

Step-by-step explanation:

Your coutour will work perfectly, so I wonder why you're hesitating to proceed with your calculation. Anyway, here is a solution:

Since sin3z=(sin3z−3sinz)/4, we have

∫∞0sin3xx3dx=18Ilimϵ→0∫R∖(−ϵ,ϵ)3eiz−e3iz−2z3dz,

where the term −2 is introduced in order to cancel out the pole of order 3, without affecting the value of the integral. Consider the counterclockwise-oriented upper semicircle C of radius R, centered at the origin, with semicircular indent of radius ϵ. Let Γ+R and γ−ϵ denote semicircular arcs of C of radius R and ϵ, respectively. If we put

f(z)=3eiz−e3iz−2z3,

then we find that

On Γ+R, we have |f(z)|≤6R−3 and thus

∫Γ+Rf(z)dz→0as R→∞.

Notice that

f(z)=3z+O(1)near z=0.

So by the direct computation,

∫γ−ϵf(z)dz=−∫π0f(ϵeiθ)iϵeiθdθ=−∫π0(3i+O(ϵ))dθ→−3πias ϵ→0.

(This is exactly the same as −πi times the residue of f at z=0. The emergence of residue can be attributed to the fact that f has only simple pole at z=0.)

Since f(z) has no pole on the region enclosed by C, we have

limϵ→0∫R∖(−ϵ,ϵ)3eiz−e3iz−2z3dz=3πi.

This proves the desired identity.

THANK YOU ❤️❤️

Answer 2

$\displaystyle \sf \red{ Let \: I=\int^{ \infty}_{0} {e}^{ - {x}^{3} } sin( {x}^{3} ) \: dx }$

$\sf \red{{e}^{ i \theta} = \cos \theta + i \sin \theta}$

$\sf \red{ \implies{e}^{ i {x}^{3} } = \cos \: x ^{3} + i \sin {x}^{3} }$

$\displaystyle \sf \red{ I = Img\bigg[\int^{ \infty}_{0} {e}^{ - {x}^{3} } \cdot {e}^{i {x}^{3} } \: dx \bigg]}$

$\displaystyle \sf \red{ \implies I = Img\bigg[\int^{ \infty}_{0} {e}^{ - {(1 - i) {x}^{3} }^{} }\: dx \bigg]}$

$\sf \red{(1 - i) {x}^{3} = t}$

$\sf \red{ \implies(1 - i) 3{x}^{2} \: dx = dt}$

$\sf \red{ \implies dx = \frac{dt}{3(1 - i) {x}^{2} } }$

$\sf \red{\because \: \frac{1}{ {x}^{2} } = \bigg(\frac{1 - i}{t} \bigg)^{ \frac{2}{3} } }$

$\sf \red{ \implies dx = \frac{dt}{3(1 - i) {}^{} } } \: \: \: \: \: \sf \red{ \frac{(1 - i)^{ \frac{2}{3} } }{ {t}^{ \frac{2}{3} }} }$

$\displaystyle \sf \red{I=Img \bigg[ \int_{0}^{ \infty} \frac{1}{3(1 - i)^{ \frac{1}{3} } } \cdot {t}^{ \frac{ - 2}{3}} {e}^{ - t} \: dt\bigg]}$

$\displaystyle \sf \red{=Img \bigg[ \frac{1}{3 [2 \sqrt{2} {e}^{ i\frac{\pi}{4} } ] ^ { \frac{1}{3} } } \int_{0}^{ \infty} {t}^{ \frac{ - 2}{3}} {e}^{ - t} \: dt\bigg]}$

$\sf{\red{ \{{ \: (1 - i) = \sqrt{2} \: {e}^{ - i \frac{\pi}{4} } \}}}}$

$\displaystyle \sf \red{ \int_{0}^{ \infty} {t}^{ \frac{ - 2}{3} } {e}^{ - t} \: dt = \Gamma \bigg( \frac{1}{3} \bigg)}$

$\sf \red{I=Img \bigg[ \frac{ {e}^{i \frac{\pi}{12} } }{3( \sqrt{2})^{ \frac{1}{3} } } \: \: \Gamma( \frac{1}{3} ) \bigg ]}$

$\sf \red{ \implies I=Img \bigg[ \frac{\Gamma( \frac{1}{3} )}{ 3 \cdot2^{ \frac{1}{6} } } \: \: ( {cos}{ \frac{\pi}{12} } + sin \frac{\pi}{12} )\bigg ]}$

$\sf \red{I = \frac{\Gamma( \frac{1}{3}) \: \: sin( \frac{\pi}{12} ) }{3 \: \cdot \: {2}^{ \frac{1}{6} } } }$