Return to Question

I needed to solve this integral:

$$\int_1^\infty \frac{dx}{\sqrt{x}} \cos \left(a x-\frac{\pi}{4} \right) \cos \left(b x-\frac{\pi}{4} \right) \cos \left(c x-\frac{\pi}{4} \right)$$

Coming from approximating the corresponding integral with Bessel functions.

The only way I saw is to expand the product into a sum of trig functions and use Fresnel integrals. With a little help from Mathematica, I've got:

$$\cos \left(a x-\frac{\pi}{4} \right) \cos \left(b x-\frac{\pi}{4} \right) \cos \left(c x-\frac{\pi}{4} \right)= \\ = \frac{1}{4} \left(\cos \left(w_1 x+\frac{\pi}{4} \right)+\cos \left(w_2 x+\frac{\pi}{4} \right)+\cos \left(w_3 x+\frac{\pi}{4} \right)-\cos \left(w_4 x+\frac{\pi}{4} \right) \right)$$

Where: $$w_1=a-b-c \\ w_2=b-a-c \\ w_3=c-a-b \\ w_4=a+b+c$$

Taking the integrals (again, with the help of Mathematica) I get:

$$\int_1^\infty \frac{dx}{\sqrt{x}} \cos \left(w x+\frac{\pi}{4} \right)= \\ =\frac{\sqrt{\pi}}{2 \sqrt{|w|}} \left(1-\operatorname{sign} w+2\operatorname{sign} w~ S \left(\frac{ \sqrt{2|w|}}{\sqrt{\pi}} \right)-2 C \left(\frac{ \sqrt{2|w|}}{\sqrt{\pi}} \right) \right)$$

This indeed agrees with the numerical integral. However, only when I set the precision of the arguments $>100$.

Can I transform the exact expression in such a way that the cancellation of the digits and the loss of precision doesn't occur? Or at least is minimal?

Here's the code (the coefficient in front of the result is slightly different than in the formula above), and the resul showing the loss of digits for WorkingPrecision->50:

a = RandomReal[{1, 100}, WorkingPrecision -> 50]; b = RandomReal[{1, 100}, WorkingPrecision -> 50]; c = RandomReal[{1, 100}, WorkingPrecision -> 50]; w1 = a - b - c; w2 = b - a - c; w3 = c - a - b; w4 = a + b + c; F[w_] := Module[{v}, v = N[Sqrt[2/\[Pi]] Sqrt[Abs[w]], 50]; N[(1 - Sign[w] - 2 FresnelC[v] + 2 Sign[w] FresnelS[v])/Sqrt[ Abs[w]], 50]]; N[{a, b, c}, 10] F[w1] + F[w2] + F[w3] - F[w4] 

Output:

{33.34824876, 95.28687965, 41.03797714} -0.011752 

For smaller precision I often don't get any significant digits.

I would really like to use this integral in an algorithm where keeping this huge precision is not very convenient, so if you have any ideas, I would be grateful.

I needed to solve this integral:

$$\int_1^\infty \frac{dx}{\sqrt{x}} \cos \left(a x-\frac{\pi}{4} \right) \cos \left(b x-\frac{\pi}{4} \right) \cos \left(c x-\frac{\pi}{4} \right)$$

Coming from approximating the corresponding integral with Bessel functions.

The only way I saw is to expand the product into a sum of trig functions and use Fresnel integrals. With a little help from Mathematica, I've got:

$$\cos \left(a x-\frac{\pi}{4} \right) \cos \left(b x-\frac{\pi}{4} \right) \cos \left(c x-\frac{\pi}{4} \right)= \\ = \frac{1}{4} \left(\cos \left(w_1 x+\frac{\pi}{4} \right)+\cos \left(w_2 x+\frac{\pi}{4} \right)+\cos \left(w_3 x+\frac{\pi}{4} \right)-\cos \left(w_4 x+\frac{\pi}{4} \right) \right)$$

Where: $$w_1=a-b-c \\ w_2=b-a-c \\ w_3=c-a-b \\ w_4=a+b+c$$

Here I use the Mathematica definition for Fresnel integrals which is different from the usual one.

Taking the integrals (again, with the help of Mathematica) I get:

$$\int_1^\infty \frac{dx}{\sqrt{x}} \cos \left(w x+\frac{\pi}{4} \right)= \\ =\frac{\sqrt{\pi}}{2 \sqrt{|w|}} \left(1-\operatorname{sign} w+2\operatorname{sign} w~ S \left(\frac{ \sqrt{2|w|}}{\sqrt{\pi}} \right)-2 C \left(\frac{ \sqrt{2|w|}}{\sqrt{\pi}} \right) \right)$$

This indeed agrees with the numerical integral. However, only when I set the precision of the arguments $>100$.

Can I transform the exact expression in such a way that the cancellation of the digits and the loss of precision doesn't occur? Or at least is minimal?

Here's the code (the coefficient in front of the result is slightly different than in the formula above), and the resul showing the loss of digits for WorkingPrecision->50:

a = RandomReal[{1, 100}, WorkingPrecision -> 50]; b = RandomReal[{1, 100}, WorkingPrecision -> 50]; c = RandomReal[{1, 100}, WorkingPrecision -> 50]; w1 = a - b - c; w2 = b - a - c; w3 = c - a - b; w4 = a + b + c; F[w_] := Module[{v}, v = N[Sqrt[2/\[Pi]] Sqrt[Abs[w]], 50]; N[(1 - Sign[w] - 2 FresnelC[v] + 2 Sign[w] FresnelS[v])/Sqrt[ Abs[w]], 50]]; N[{a, b, c}, 10] F[w1] + F[w2] + F[w3] - F[w4] 

Output:

{33.34824876, 95.28687965, 41.03797714} -0.011752 

For smaller precision I often don't get any significant digits.

I would really like to use this integral in an algorithm where keeping this huge precision is not very convenient, so if you have any ideas, I would be grateful.

I needed to solve this integral:

$$\int_1^\infty \frac{dx}{\sqrt{x}} \cos \left(a x-\frac{\pi}{4} \right) \cos \left(b x-\frac{\pi}{4} \right) \cos \left(c x-\frac{\pi}{4} \right)$$

Coming from approximating the corresponding integral with Bessel functions.

The only way I saw is to expand the product into a sum of trig functions and use Fresnel integrals. With a little help from Mathematica, I've got:

$$\cos \left(a x-\frac{\pi}{4} \right) \cos \left(b x-\frac{\pi}{4} \right) \cos \left(c x-\frac{\pi}{4} \right)= \\ = \frac{1}{4} \left(\cos \left(w_1 x+\frac{\pi}{4} \right)+\cos \left(w_2 x+\frac{\pi}{4} \right)+\cos \left(w_3 x+\frac{\pi}{4} \right)-\cos \left(w_4 x+\frac{\pi}{4} \right) \right)$$

Where: $$w_1=a-b-c \\ w_2=b-a-c \\ w_3=c-a-b \\ w_4=a+b+c$$

Taking the integrals (again, with the help of Mathematica) I get:

$$\int_1^\infty \frac{dx}{\sqrt{x}} \cos \left(w x+\frac{\pi}{4} \right)= \\ =\frac{\sqrt{\pi}}{2 \sqrt{|w|}} \left(1-\operatorname{sign} w+2\operatorname{sign} w~ S \left(\frac{2 \sqrt{|w|}}{\sqrt{\pi}} \right)-2 C \left(\frac{2 \sqrt{|w|}}{\sqrt{\pi}} \right) \right)$$

This indeed agrees with the numerical integral. However, only when I set the precision of the arguments $>100$.

Can I transform the exact expression in such a way that the cancellation of the digits and the loss of precision doesn't occur? Or at least is minimal?

Here's the code, and the resul showing the loss of digits for WorkingPrecision->50:

a = RandomReal[{1, 100}, WorkingPrecision -> 50]; b = RandomReal[{1, 100}, WorkingPrecision -> 50]; c = RandomReal[{1, 100}, WorkingPrecision -> 50]; w1 = a - b - c; w2 = b - a - c; w3 = c - a - b; w4 = a + b + c; F[w_] := Module[{v}, v = N[Sqrt[2/\[Pi]] Sqrt[Abs[w]], 50]; N[(1 - Sign[w] - 2 FresnelC[v] + 2 Sign[w] FresnelS[v])/Sqrt[ Abs[w]], 50]]; N[{a, b, c}, 10] F[w1] + F[w2] + F[w3] - F[w4] 

Output:

{33.34824876, 95.28687965, 41.03797714} -0.011752 

For smaller precision I often don't get any significant digits.

I would really like to use this integral in an algorithm where keeping this huge precision is not very convenient, so if you have any ideas, I would be grateful.

I needed to solve this integral:

$$\int_1^\infty \frac{dx}{\sqrt{x}} \cos \left(a x-\frac{\pi}{4} \right) \cos \left(b x-\frac{\pi}{4} \right) \cos \left(c x-\frac{\pi}{4} \right)$$

Coming from approximating the corresponding integral with Bessel functions.

The only way I saw is to expand the product into a sum of trig functions and use Fresnel integrals. With a little help from Mathematica, I've got:

$$\cos \left(a x-\frac{\pi}{4} \right) \cos \left(b x-\frac{\pi}{4} \right) \cos \left(c x-\frac{\pi}{4} \right)= \\ = \frac{1}{4} \left(\cos \left(w_1 x+\frac{\pi}{4} \right)+\cos \left(w_2 x+\frac{\pi}{4} \right)+\cos \left(w_3 x+\frac{\pi}{4} \right)-\cos \left(w_4 x+\frac{\pi}{4} \right) \right)$$

Where: $$w_1=a-b-c \\ w_2=b-a-c \\ w_3=c-a-b \\ w_4=a+b+c$$

Taking the integrals (again, with the help of Mathematica) I get:

$$\int_1^\infty \frac{dx}{\sqrt{x}} \cos \left(w x+\frac{\pi}{4} \right)= \\ =\frac{\sqrt{\pi}}{2 \sqrt{|w|}} \left(1-\operatorname{sign} w+2\operatorname{sign} w~ S \left(\frac{ \sqrt{2|w|}}{\sqrt{\pi}} \right)-2 C \left(\frac{ \sqrt{2|w|}}{\sqrt{\pi}} \right) \right)$$

This indeed agrees with the numerical integral. However, only when I set the precision of the arguments $>100$.

Can I transform the exact expression in such a way that the cancellation of the digits and the loss of precision doesn't occur? Or at least is minimal?

Here's the code (the coefficient in front of the result is slightly different than in the formula above), and the resul showing the loss of digits for WorkingPrecision->50:

a = RandomReal[{1, 100}, WorkingPrecision -> 50]; b = RandomReal[{1, 100}, WorkingPrecision -> 50]; c = RandomReal[{1, 100}, WorkingPrecision -> 50]; w1 = a - b - c; w2 = b - a - c; w3 = c - a - b; w4 = a + b + c; F[w_] := Module[{v}, v = N[Sqrt[2/\[Pi]] Sqrt[Abs[w]], 50]; N[(1 - Sign[w] - 2 FresnelC[v] + 2 Sign[w] FresnelS[v])/Sqrt[ Abs[w]], 50]]; N[{a, b, c}, 10] F[w1] + F[w2] + F[w3] - F[w4] 

Output:

{33.34824876, 95.28687965, 41.03797714} -0.011752 

For smaller precision I often don't get any significant digits.

I would really like to use this integral in an algorithm where keeping this huge precision is not very convenient, so if you have any ideas, I would be grateful.