Monday, 29 July 2013

integration - Definite integral involving Fresnel integrals




I am seeking to evaluate



0f(x)/x2dx



with



f(x)=1π/6(cos(x)C(6xπ)+S(6xπ)sin(x))/x.



C(x) and S(x) are the Fresnel integrals. Numerical integration suggests that the integral equals 2π/(33), which would also be desirable within the (physical) context it arose. How can this be proved?



Answer



First of all,
C(x)=x0cos(12πt2)dt=2ππ/2(x)0cos(z2)dz,
and
S(x)=x0sin(12πt2)dt=2ππ/2(x)0sin(z2)dz.
So we have

C(6xπ)=2π3x0cos(z2)dz,
and
S(6xπ)=2π3x0sin(z2)dz.
Using these we can write
f(x)=1π/6cos(x)C(6xπ)+sin(x)S(6xπ)x=3x0(1cos(xz2))dz3x
and
0f(x)x2dx=2303x0sin2((xz2)/2)x5/2dzdx.
Let's introduce the new variable z=tx. Then dz=xdt and
03x0sin2((xz2)/2)x5/2dzdx=030sin2(x(1t2)/2)x2dtdx=12030sin2(x(1t2))x2dtdx=12010sin2(x(1t2))x2dtdx+12031sin2(x(t21))x2dtdx=12100sin2(x(1t2))x2dxdt+12310sin2(x(t21))x2dxdt=10π4(1t2)dt+31π4(t21)dt=π3,

where the formula
0sin2(Ax)x2dx=12Aπ,(A>0)
was applied. Your conjecture was excellent.


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