I am seeking to evaluate
∫∞0f(x)/x2dx
with
f(x)=1−√π/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/(3√3), 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=√2π∫√π/2(x)0cos(z2)dz,
and
S(x)=∫x0sin(12πt2)dt=√2π∫√π/2(x)0sin(z2)dz.
So we have
C(√6xπ)=√2π∫√3x0cos(z2)dz,
and
S(√6xπ)=√2π∫√3x0sin(z2)dz.
Using these we can write
f(x)=1−√π/6cos(x)C(√6xπ)+sin(x)S(√6xπ)√x=∫√3x0(1−cos(x−z2))dz√3x
and
∫∞0f(x)x2dx=2√3∫∞0∫√3x0sin2((x−z2)/2)x5/2dzdx.
Let's introduce the new variable z=t√x. Then dz=√xdt and
∫∞0∫√3x0sin2((x−z2)/2)x5/2dzdx=∫∞0∫√30sin2(x(1−t2)/2)x2dtdx=12∫∞0∫√30sin2(x(1−t2))x2dtdx=12∫∞0∫10sin2(x(1−t2))x2dtdx+12∫∞0∫√31sin2(x(t2−1))x2dtdx=12∫10∫∞0sin2(x(1−t2))x2dxdt+12∫√31∫∞0sin2(x(t2−1))x2dxdt=∫10π4(1−t2)dt+∫√31π4(t2−1)dt=π3,
where the formula
∫∞0sin2(Ax)x2dx=12Aπ,(A>0)
was applied. Your conjecture was excellent.
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