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)e−λtdt=λα2+λ2

We will also use the standard arctangent integral:

∫∞0dta2+t2=π2a

Now

(∫∞0sin(x2)e−λx2dx)2=∫∞0∫∞0sin(x2)sin(y2)e−λ(x2+y2)dydx=12∫∞0∫∞0(cos(x2−y2)−cos(x2+y2))e−λ(x2+y2)dydx=12∫π/20∫∞0(cos(r2cos(2ϕ))−cos(r2))e−λr2rdrdϕ=14∫π/20∫∞0(cos(scos(2ϕ))−cos(s))e−λsdsdϕ=14∫π/20(λcos2(2ϕ)+λ2−λ1+λ2)dϕ=12∫π/40λcos2(2ϕ)+λ2dϕ−λπ/81+λ2=14∫π/40λdtan(2ϕ)1+λ2+λ2tan2(2ϕ)−λπ/81+λ2=14∫∞0dt1+λ2+t2−λπ/81+λ2=π/8√1+λ2−λπ/81+λ2

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

(3.2) use 2sin(A)sin(B)=cos(A−B)−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 λ→0+ 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*