Evaluating ∫∞0sinx2dx\int_0^\infty \sin x^2\, dx with real methods?

I have seen the Fresnel integral

0sinx2dx=π8

evaluated by contour integration and other complex analysis methods, and I have found these methods to be the standard way to evaluate this integral. I was wondering, however, does anyone know a real analysis method to evaluate this integral?

Answer

Aside from some trigonometric substitutions and identities, we will need the following identity, which can be shown using integration by parts twice:
0cos(αt)eλtdt=λα2+λ2
We will also use the standard arctangent integral:
0dta2+t2=π2a
Now
(0sin(x2)eλx2dx)2=00sin(x2)sin(y2)eλ(x2+y2)dydx=1200(cos(x2y2)cos(x2+y2))eλ(x2+y2)dydx=12π/200(cos(r2cos(2ϕ))cos(r2))eλr2rdrdϕ=14π/200(cos(scos(2ϕ))cos(s))eλsdsdϕ=14π/20(λcos2(2ϕ)+λ2λ1+λ2)dϕ=12π/40λcos2(2ϕ)+λ2dϕλπ/81+λ2=14π/40λdtan(2ϕ)1+λ2+λ2tan2(2ϕ)λπ/81+λ2=140dt1+λ2+t2λπ/81+λ2=π/81+λ2λπ/81+λ2

(3.1) change the square of the integral into a double integral

(3.2) use 2sin(A)sin(B)=cos(AB)cos(A+B)

(3.3) convert to polar coordinates

(3.4) substitute s=r2

(3.5) apply (1)

(3.6) pull out the constant and apply symmetry to reduce the domain of integration

(3.7) multiply numerator and denominator by sec2(2ϕ)

(3.8) substitute t=λtan(2ϕ)

(3.9) apply (2)

Finally, take the square root of both sides of (3) and let λ0+ to get
0sin(x2)dx=π8

Addendum

I just noticed that the same proof works for
0cos(x2)dx=π8
if each red sin is changed to cos and each red sign is changed to +.

Attribution
Source : Link , Question Author : Argon , Answer Author : robjohn

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