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# 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.

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

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

Therefore, we get:

$$\displaystyle LHS=\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ { x }^{ \frac { -3 }{ 4 } }cos\left( { a }\sqrt { x } \right) dx }$$

Now by Maclaurin series,

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

This can also be written as:

$$\displaystyle cos\left( { a }\sqrt { x } \right) =\sum _{ n=0 }^{ \infty }{ \frac { { \left( -x \right) }^{ n }{ \left( { a } \right) }^{ 2n }n! }{ 2n!n! } }$$

On plugging the value into LHS, we get:

$$\displaystyle LHS=\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ { x }^{ \frac { -3 }{ 4 } }\sum _{ n=0 }^{ \infty }{ \frac { { \left( -x \right) }^{ n }{ \left( { a } \right) }^{ 2n }n! }{ 2n!n! } } dx }$$

Now, by Ramanujan's Master Theorem, we get

$$\displaystyle LHS=\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ { x }^{ \frac { -3 }{ 4 } }\sum _{ n=0 }^{ \infty }{ \frac { { \left( -x \right) }^{ n }{ \left( { a } \right) }^{ 2n }n! }{ 2n!n! } } dx } =\frac { 1 }{ 4 } \frac { \Gamma \left( \frac { 1 }{ 4 } \right) \Gamma \left( \frac { 3 }{ 4 } \right) }{ \sqrt { a } \Gamma \left( \frac { 1 }{ 2 } \right) }$$

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

Now, we consider RHS as:

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

On differentiating both the sides wrt a, we get:

$$\displaystyle I'\left( a \right) =\int _{ 0 }^{ \infty }{ { x }^{ 2 }cos\left( { ax }^{ 2 } \right) dx }$$

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

Therefore, we get

$$\displaystyle I'\left( a \right) =\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ { x }^{ \frac { -1 }{ 4 } }cos\left( { a }\sqrt { x } \right) dx }$$

On evaluating this integral as above, we get

$$\displaystyle I'\left( a \right) =\frac { 1 }{ 4 } \frac { \Gamma \left( \frac { 1 }{ 4 } \right) \Gamma \left( \frac { 3 }{ 4 } \right) }{ \sqrt { { a }^{ 3 } } \Gamma \left( \frac { -1 }{ 2 } \right) }$$

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

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

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

$\text{HENCE PROVED}$

#### ORIGINAL

12 months ago

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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 $\int_0^\infty \frac{e^{-xt}}{\sqrt{t}}\,dt \; = \; 2\int_0^\infty e^{-xs^2}\,ds \; = \; \frac{2}{\sqrt{x}}\int_0^\infty e^{-s^2}\,ds \; = \; \sqrt{\frac{\pi}{x}}$ for $$x > 0$$, and so $\begin{array}{rcl} \displaystyle \int_0^R \sin y^2\,dy & = & \displaystyle \frac12\int_0^{R^2} \frac{\sin x}{\sqrt{x}}\,dx \; = \; \frac12\int_0^{R^2} \sin x \left(\int_0^\infty \frac{e^{-xt}}{\sqrt{\pi t}}\,dt\right)\,dx \\ & = & \displaystyle \frac{1}{2\sqrt{\pi}}\int_0^\infty \frac{1}{\sqrt{t}}\left(\int_0^{R^2} e^{-xt} \sin x\,dx\right) \\ & = & \displaystyle \frac{1}{2\sqrt{\pi}}\int_0^\infty \frac{1}{\sqrt{t}(1+t^2)}\big[1 - e^{-R^2t}\cos R^2 - te^{-R^2t}\sin R^2\big]\,dt \end{array}$ All these integrals exist in a Lebesgue sense, and Fubini/Tonelli justifies reversing the order of integration. Since $$\frac{1+t}{\sqrt{t}(t^2+1)}$$ is Lebesgue-integrable on $$(0,\infty)$$, the Dominated Convergence Theorem enables us to let $$R \to \infty$$ in the above, obtaining the improper Riemann integral $\int_0^\infty \sin y^2\,dy \; = \; \lim_{R\to\infty} \int_0^R \sin y^2\,dy \; = \; \frac{1}{2\sqrt{\pi}} \int_0^\infty \frac{1}{\sqrt{t}(1+t^2)}\,dt \; = \; \frac{1}{\sqrt{\pi}}\int_0^\infty \frac{1}{1+s^4}\,ds \; = \; \frac{\sqrt{\pi}}{2\sqrt{2}} \;.$ A similar argument works for the cosine. Adding the positive term $$a$$ is achieved by a trivial change of variable. · 12 months ago

Ooh so I needed to prove that it coverages when R tends to infinity. · 12 months ago

Yes. What everyone means (but rarely makes explicit) when they write $\int_0^\infty \sin y^2\,dy$ is the limit of the integral from $$0$$ to $$R$$. 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 $$\sin y^2$$ is not Lebesgue integrable on $$(0,\infty)$$, and that is where the difficulties come in... · 12 months ago

I'll make sure to prove it next time onwards. Thanks! · 12 months ago

Contour integration is another method. :D for proving this · 11 months, 3 weeks ago

Indeed. Integrating around the "octant" contour comprising

• the straight line from $$0$$ to $$R$$,
• the circular arc $$z = Re^{I\theta}$$ for $$0 \le \theta \le \tfrac14\pi$$,
• the straight line from $$\tfrac1{\sqrt{2}}R(1+i)$$ back to $$0$$

and letting $$R \to \infty$$ gives us $\lim_{R\to\infty} \int_0^R e^{ix^2}\,dx \; = \; \frac{1+i}{\sqrt{2}}\int_0^\infty e^{-x^2}\,dx \; = \; \frac{(1+i)\pi}{2\sqrt{2}}$ which does the trick. Showing that the integral around the circular arc tends to zero is not straightforward, though. If $$z = Re^{i\theta}$$ then $$\big|e^{iz^2}\big| \,=\, e^{-R^2\sin2\theta} \; \le \; e^{-\frac{4}{\pi}R^2\theta}$$ for $$0 \le \theta \le \tfrac14\pi$$, and this makes the integral along the curved arc $$O(R^{-1})$$. · 11 months, 3 weeks ago