Proof of Fresnel Integrals

We have to prove: \[\int _{ 0 }^{ \infty }{ \cos\left( { ax }^{ 2 } \right) \, dx } =\sqrt { \frac { \pi }{ 8a } } =\int _{ 0 }^{ \infty }{ \sin\left( { ax }^{ 2 } \right) \, dx } \]

Now, we consider the LHS.

LHS=∫0∞cos(ax2)dx\displaystyle LHS=\int _{ 0 }^{ \infty }{ cos\left( { ax }^{ 2 } \right) dx }

Now, we make the substitution: x→x14x\rightarrow { x }^{ \frac { 1 }{ 4 } }

Therefore, we get:

LHS=14∫0∞x−34cos(ax)dx\displaystyle LHS=\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ { x }^{ \frac { -3 }{ 4 } }cos\left( { a }\sqrt { x } \right) dx }

Now by Maclaurin series,

cos(ax)=∑n=0∞(−1)n(ax)2n2n!\displaystyle cos\left( { a }\sqrt { x } \right) =\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n }{ \left( { a }\sqrt { x } \right) }^{ 2n } }{ 2n! } }

This can also be written as:

cos(ax)=∑n=0∞(−x)n(a)2nn!2n!n!\displaystyle cos\left( { a }\sqrt { x } \right) =\sum _{ n=0 }^{ \infty }{ \frac { { \left( -x \right) }^{ n }{ \left( { a } \right) }^{ 2n }n! }{ 2n!n! } }

On plugging the value into LHS, we get:

LHS=14∫0∞x−34∑n=0∞(−x)n(a)2nn!2n!n!dx\displaystyle LHS=\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ { x }^{ \frac { -3 }{ 4 } }\sum _{ n=0 }^{ \infty }{ \frac { { \left( -x \right) }^{ n }{ \left( { a } \right) }^{ 2n }n! }{ 2n!n! } } dx }

Now, by Ramanujan's Master Theorem, we get

LHS=14∫0∞x−34∑n=0∞(−x)n(a)2nn!2n!n!dx=14Γ(14)Γ(34)aΓ(12)\displaystyle LHS=\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ { x }^{ \frac { -3 }{ 4 } }\sum _{ n=0 }^{ \infty }{ \frac { { \left( -x \right) }^{ n }{ \left( { a } \right) }^{ 2n }n! }{ 2n!n! } } dx } =\frac { 1 }{ 4 } \frac { \Gamma \left( \frac { 1 }{ 4 } \right) \Gamma \left( \frac { 3 }{ 4 } \right) }{ \sqrt { a } \Gamma \left( \frac { 1 }{ 2 } \right) }

Therefore, by Euler's Reflection Formula, we get ∫0∞cos(ax2)dx=π8a\int _{ 0 }^{ \infty }{ cos\left( { ax }^{ 2 } \right) dx } =\sqrt { \frac { \pi }{ 8a } }

Now, we consider RHS as:

I(a)=∫0∞sin(ax2)dx\displaystyle I\left( a \right) =\int _{ 0 }^{ \infty }{ sin\left( { ax }^{ 2 } \right) dx }

On differentiating both the sides wrt a, we get:

I′(a)=∫0∞x2cos(ax2)dx\displaystyle I'\left( a \right) =\int _{ 0 }^{ \infty }{ { x }^{ 2 }cos\left( { ax }^{ 2 } \right) dx }

Now we make the substitution: x→x14x\rightarrow { x }^{ \frac { 1 }{ 4 } }

Therefore, we get

I′(a)=14∫0∞x−14cos(ax)dx\displaystyle I'\left( a \right) =\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ { x }^{ \frac { -1 }{ 4 } }cos\left( { a }\sqrt { x } \right) dx }

On evaluating this integral as above, we get

I′(a)=14Γ(14)Γ(34)a3Γ(−12)\displaystyle I'\left( a \right) =\frac { 1 }{ 4 } \frac { \Gamma \left( \frac { 1 }{ 4 } \right) \Gamma \left( \frac { 3 }{ 4 } \right) }{ \sqrt { { a }^{ 3 } } \Gamma \left( \frac { -1 }{ 2 } \right) }

Since I(0)=0I\left( 0 \right) =0 we can directly integrate the above expression wrt a.

Therefore we get:∫0∞sin(ax2)dx=π8a\int _{ 0 }^{ \infty }{ sin\left( { ax }^{ 2 } \right) dx } =\sqrt { \frac { \pi }{ 8a } }

Hence we can conclude that ∫0∞cos(ax2)dx=π8a=∫0∞sin(ax2)dx\int _{ 0 }^{ \infty }{ cos\left( { ax }^{ 2 } \right) dx } =\sqrt { \frac { \pi }{ 8a } } =\int _{ 0 }^{ \infty }{ sin\left( { ax }^{ 2 } \right) dx } .

HENCE PROVED\text{HENCE PROVED}

ORIGINAL

#Calculus

Note by Aditya Kumar
5 years, 4 months ago

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Comments

As I discussed with you previously concerning the Ramanujan Master Theorem, the RMT is only a formal argument (albeit one that frequently gives good results) except when the integral it is trying to evaluate is properly convergent. This is another case where this is not true. Now ∫0∞e−xtt dt  =  2∫0∞e−xs2 ds  =  2x∫0∞e−s2 ds  =  πx \int_0^\infty \frac{e^{-xt}}{\sqrt{t}}\,dt \; = \; 2\int_0^\infty e^{-xs^2}\,ds \; = \; \frac{2}{\sqrt{x}}\int_0^\infty e^{-s^2}\,ds \; = \; \sqrt{\frac{\pi}{x}} for x>0x > 0, and so ∫0Rsin⁡y2 dy=12∫0R2sin⁡xx dx  =  12∫0R2sin⁡x(∫0∞e−xtπt dt) dx=12π∫0∞1t(∫0R2e−xtsin⁡x dx)=12π∫0∞1t(1+t2)[1−e−R2tcos⁡R2−te−R2tsin⁡R2] dt \begin{array}{rcl} \displaystyle \int_0^R \sin y^2\,dy & = & \displaystyle \frac12\int_0^{R^2} \frac{\sin x}{\sqrt{x}}\,dx \; = \; \frac12\int_0^{R^2} \sin x \left(\int_0^\infty \frac{e^{-xt}}{\sqrt{\pi t}}\,dt\right)\,dx \\ & = & \displaystyle \frac{1}{2\sqrt{\pi}}\int_0^\infty \frac{1}{\sqrt{t}}\left(\int_0^{R^2} e^{-xt} \sin x\,dx\right) \\ & = & \displaystyle \frac{1}{2\sqrt{\pi}}\int_0^\infty \frac{1}{\sqrt{t}(1+t^2)}\big[1 - e^{-R^2t}\cos R^2 - te^{-R^2t}\sin R^2\big]\,dt \end{array} All these integrals exist in a Lebesgue sense, and Fubini/Tonelli justifies reversing the order of integration. Since 1+tt(t2+1)\frac{1+t}{\sqrt{t}(t^2+1)} is Lebesgue-integrable on (0,∞)(0,\infty), the Dominated Convergence Theorem enables us to let R→∞R \to \infty in the above, obtaining the improper Riemann integral ∫0∞sin⁡y2 dy  =  lim⁡R→∞∫0Rsin⁡y2 dy  =  12π∫0∞1t(1+t2) dt  =  1π∫0∞11+s4 ds  =  π22  . \int_0^\infty \sin y^2\,dy \; = \; \lim_{R\to\infty} \int_0^R \sin y^2\,dy \; = \; \frac{1}{2\sqrt{\pi}} \int_0^\infty \frac{1}{\sqrt{t}(1+t^2)}\,dt \; = \; \frac{1}{\sqrt{\pi}}\int_0^\infty \frac{1}{1+s^4}\,ds \; = \; \frac{\sqrt{\pi}}{2\sqrt{2}} \;. A similar argument works for the cosine. Adding the positive term aa is achieved by a trivial change of variable.

Mark Hennings - 5 years, 4 months ago

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Ooh so I needed to prove that it coverages when R tends to infinity.

Aditya Kumar - 5 years, 4 months ago

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Yes. What everyone means (but rarely makes explicit) when they write ∫0∞sin⁡y2 dy \int_0^\infty \sin y^2\,dy is the limit of the integral from 00 to RR. That is the only way that Riemann integration can define integrals to infinity. Lebesgue integration can define integrals on infinite integrals as easily as it can on finite ones. The function sin⁡y2\sin y^2 is not Lebesgue integrable on (0,∞)(0,\infty), and that is where the difficulties come in...

Mark Hennings - 5 years, 4 months ago

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@Mark Hennings – I'll make sure to prove it next time onwards. Thanks!

Aditya Kumar - 5 years, 4 months ago

Contour integration is another method. :D for proving this

Aman Rajput - 5 years, 4 months ago

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Indeed. Integrating around the "octant" contour comprising

  • the straight line from 00 to RR,
  • the circular arc z=ReIθz = Re^{I\theta} for 0≤θ≤14π0 \le \theta \le \tfrac14\pi,
  • the straight line from 12R(1+i)\tfrac1{\sqrt{2}}R(1+i) back to 00

and letting R→∞R \to \infty gives us lim⁡R→∞∫0Reix2 dx  =  1+i2∫0∞e−x2 dx  =  (1+i)π22 \lim_{R\to\infty} \int_0^R e^{ix^2}\,dx \; = \; \frac{1+i}{\sqrt{2}}\int_0^\infty e^{-x^2}\,dx \; = \; \frac{(1+i)\pi}{2\sqrt{2}} which does the trick. Showing that the integral around the circular arc tends to zero is not straightforward, though. If z=Reiθz = Re^{i\theta} then ∣eiz2∣ = e−R2sin⁡2θ  ≤  e−4πR2θ\big|e^{iz^2}\big| \,=\, e^{-R^2\sin2\theta} \; \le \; e^{-\frac{4}{\pi}R^2\theta} for 0≤θ≤14π0 \le \theta \le \tfrac14\pi, and this makes the integral along the curved arc O(R−1)O(R^{-1}).

Mark Hennings - 5 years, 4 months ago

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yes , correct !

Aman Rajput - 5 years, 4 months ago

Please explain me the step after RMT and before Euler’s reflection formula

Abhishek Keripale - 3 years, 8 months ago
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