# A Double Integral Containing the Fresnel Integral FunctionSx: Derivation and Computation

Authors: Robert Reynolds; Allan Stauffer
Journal: Journal of Mathematics
 (2022)

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## Abstract

A two-dimensional integral containingSx is derived. Sx is the Fresnel integral function, and the double integral is taken over the range 0<x<∞and0<y<∞. A representation in terms of the Hurwitz–Lerch zeta function is derived, from which other special function representations can be evaluated. All the results in this work are new.


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## Body

## 1. Significance Statement

The Fresnel integralSx named after Augustin-Jean Fresnel (1788–1827) was first studied by Euler in 1781. Fresnel used this function in 1818 to study the illumination intensity of optical diffraction patterns [1]. The Fresnel integral Sx is an odd function and has mirror symmetry [2, 3].Knochenhauer [4] derived series representations of the Fresnel integrals. Leipzig and Teubner [5] published a number of properties of the Fresnel integrals. Liou [6] used the double integral containing the product of Fresnel integrals to investigate the Lorenz–Mie theory of light scattering by spherical particles. In their work, the electromagnetic wave equation was studied and a solution was provided. Fraunhofer diffraction [7], asymptotics of Weyl sums [8], and railway and road projects [9] are all examples of Fresnel integral applications.Indefinite and definite integrals whose kernels feature the Fresnel integral are tabled in the book of Prudnikov et al. [10]. Multiple integrals of Fresnel integrals is of high importance based on the current published work. In our present paper, we aim to expand the current work featuring the double integral of the Fresnel integral by providing additional integral formulae to aid in research requiring such integrals.


## 2. Introduction

In this paper, we derive the double definite integral given by(1)∫0∞∫0∞α3/2e−by2xm+1/2y−2mS2/πxαlogk−1ax/y2mlogax/y2+kα3x3dxdy,where the hypergeometric function representation for the Fresnel integral Sx is given in Table (3.7) in [11] and parameters k,a,andm are general complex numbers and Reb>0andRem<−1/2<Reα. This definite integral will be used to derive special cases in terms of special functions and fundamental constants. The derivations follow the method used by us in [12, 13]. This method involves using a form of the generalized Cauchy’s integral formula given by(2)ykΓk+1=12πi∫Cewywk+1dw,where C is, in general, an open contour in the complex plane where the bilinear concomitant has the same value at the end points of the contour. We then multiply both sides by a function of x and y and then take a definite double integral of both sides. This yields a definite integral in terms of a contour integral. Then, we multiply both sides of equation (2) by another function of y and take the infinite sum of both sides such that the contour integral of both equations are the same.


## 3. Definite Integral of the Contour Integral

We used the method in [12]. The variable of integration in the contour integral is r=w+m. The cut and contour are in the first quadrant of the complex r-plane. The cut approaches the origin from the interior of the first quadrant; the contour goes round the origin with zero radius and is on opposite sides of the cut. Using a generalization of Cauchy’s integral formula, we form the double integral by forming two equations by first replacing k by k−1 and y with(3)logaxy2,and multiplying by(4)e−by2xm−5/2y−2mα3x3S2/πxαα3/2.For the first equation, replacey by(5)logaxy2,leave k, and multiply by(6)me−by2xm−5/2y−2mα3x3S2/πxαα3/2.For the second equation, then add both equations taking the definite integral with respect tox∈0,∞ and y∈0,∞ to obtain(7)1Γk+1∫0∞∫0∞α3/2e−by2xm+1/2y−2mS2/πxαlogk−1ax/y2mlogax/y2+kα3x3dxdy=12πi∫0∞∫0∞∫Cawe−by2w−k−1m+wSxαxm+w−1y−2m+wdwdxdy=12πi∫C∫0∞∫0∞awe−by2w−k−1m+wSxαxm+w−1y−2m+wdxdydw=−12πi∫C12π2aww−k−1bm+w−1/2α−m−wsin14π2m+w+1secπm+wdw.The above expression is simplified using equations (3.7.1) in [11], equation (3.326.2) in [14] where Rem+w>0, and the reflection formula (8.334.3) in [14] for the Gamma function. We are able to switch the order of integration over x, y, and r using Fubini’s theorem for multiple integrals (see (9.112) in [15]), since the integrand is of bounded measure over the space ℂ×0,∞×0,∞.


## 4. The Hurwitz–Lerch Zeta Function and Infinite Sum of the Contour Integral

In this section, we use equation (2) to derive the contour integral representations for the Hurwitz–Lerch Zeta function.


### 4.1. The Hurwitz–Lerch Zeta Function

The Hurwitz–Lerch Zeta function in [16–18] has a series representation given by(8)Φz,s,v=∑n=0∞v+n−szn,where z<1,v≠0,−1,… and is continued analytically by its integral representation given by(9)Φz,s,v=1Γs∫0∞ts−1e−vt1−ze−tdt=1Γs∫0∞ts−1e−v−1tet−zdt,where Rev>0, and when either z≤1andz≠1,Res>0, or z=1,Res>1.


### 4.2. Infinite Sum of the Contour Integral

Use equation (2) and replace y by(10)y+2ix.Then, multiply both sides by(11)eimx.Next, we form a second equation by replacingx⟶−x and taking their difference and replace x⟶π/4andm⟶2m+1 to get(12)e−1/4iπ2m+1y−iπ/2k−e1/2iπ2m+1y+iπ/2kΓk+1−12πi∫C2iw−k−1ewysin14π2m+2w+1dw.Next, using equation (12), we replace y by(13)loga−logα+logb+iπ2y+1,and multiply both sides by(14)−12iπ2bm−1/2α−m−1yeiπm2y+1.Then, taking the infinite sum over y∈0,∞ and simplifying in terms of the Hurwitz–Lerch Zeta function, we obtain(15)−1+i2k−2πk+1/2bm−1/2e1/2iπk+mα−mΓk+1Φ−e2imπ,−k,−2iloga−2ilogb+2ilogα+π4π−ieiπmΦ−e2imπ,−k,−2iloga−2ilogb+2ilogα+3π4π=−12πi∑y=0∞∫C2i−1yw−k−1sin14π2m+2w+1expwloga−logα+logb+iπ2y+1m+wdw=−12πi∫C2i−1yw−k−1sin14π2m+2w+1⋅∑y=0∞expwloga−logα+logb+iπ2y+1m+wdw=−12πi∫C12π2aww−k−1bm+w−1/2α−m−wsin14π2m+2w+1secπm+wdw,from Equation (1.232.3) in [14], where Imm+w>0 in order for the sum to converge.


## 4.1. The Hurwitz–Lerch Zeta Function

The Hurwitz–Lerch Zeta function in [16–18] has a series representation given by(8)Φz,s,v=∑n=0∞v+n−szn,where z<1,v≠0,−1,… and is continued analytically by its integral representation given by(9)Φz,s,v=1Γs∫0∞ts−1e−vt1−ze−tdt=1Γs∫0∞ts−1e−v−1tet−zdt,where Rev>0, and when either z≤1andz≠1,Res>0, or z=1,Res>1.


## 4.2. Infinite Sum of the Contour Integral

Use equation (2) and replace y by(10)y+2ix.Then, multiply both sides by(11)eimx.Next, we form a second equation by replacingx⟶−x and taking their difference and replace x⟶π/4andm⟶2m+1 to get(12)e−1/4iπ2m+1y−iπ/2k−e1/2iπ2m+1y+iπ/2kΓk+1−12πi∫C2iw−k−1ewysin14π2m+2w+1dw.Next, using equation (12), we replace y by(13)loga−logα+logb+iπ2y+1,and multiply both sides by(14)−12iπ2bm−1/2α−m−1yeiπm2y+1.Then, taking the infinite sum over y∈0,∞ and simplifying in terms of the Hurwitz–Lerch Zeta function, we obtain(15)−1+i2k−2πk+1/2bm−1/2e1/2iπk+mα−mΓk+1Φ−e2imπ,−k,−2iloga−2ilogb+2ilogα+π4π−ieiπmΦ−e2imπ,−k,−2iloga−2ilogb+2ilogα+3π4π=−12πi∑y=0∞∫C2i−1yw−k−1sin14π2m+2w+1expwloga−logα+logb+iπ2y+1m+wdw=−12πi∫C2i−1yw−k−1sin14π2m+2w+1⋅∑y=0∞expwloga−logα+logb+iπ2y+1m+wdw=−12πi∫C12π2aww−k−1bm+w−1/2α−m−wsin14π2m+2w+1secπm+wdw,from Equation (1.232.3) in [14], where Imm+w>0 in order for the sum to converge.


## 5. Definite Integral in terms of the Hurwitz–Lerch Zeta Function

Theorem 1.
For allk,a,m∈ℂ,Reb>0,Reα>0,andRem<−1/2<Reα, then(16)∫0∞∫0∞α3/2e−by2xm+1/2y−2mS2/πxαlogk−1ax/y2mlogax/y2+kα3x3dxdy=−1−i2k−2πk+1/2bm−1/2e1/2iπk+mα−mΦ−e2imπ,−k,−2iloga−2ilogb+2ilogα+π4π−ieiπmΦ−e2imπ,−k,−2iloga−2ilogb+2ilogα+3π4π.Proof.
The right-hand sides of relations (7) and (15) are identical; hence, the left-hand sides of the same are identical too. Simplifying with the Gamma function yields the desired conclusion.Example 1.
The degenerate case.(17)∫0∞∫0∞α3/2me−by2xm+1/2y−2mS2/πxαα3x3dxdy=−14π2bm−1/2α−msec142πm+π.Proof.
Use Equation (16), set k=0, and simplify using entry (2) in the table below equation (64 : 12 : 7) in [19].Example 2.
The Polylogarithm functionLikz.(18)∫0∞∫0∞x3e−y2S2πxxm−5/2y−2mlogk−1ixy2k+mlogixy2dxdy=−1−i2k−2πk+1/2e1/2iπk+mΦ−e2imπ,−k,12+ie−iπmLi−k−e2imπ.Proof.
Use equation (16), set α=1,b=1,a=i, and simplify using equation (64 : 12 : 2) in [19].Example 3.
Catalan’s constantK.(19)∫0∞∫0∞x3e−y2y3/2S2/πx8+3logix/y2x13/4log3ix/y2dxdy=−15/8−48K2+48+48iΦ−i,2,1/2+i2π2192π3/2.Proof.
Use equation (18), set k=−2,m=−3/4, and simplify using equation (2.2.1.2.7) in [20].Example 4.
The Hurwitz zeta functionζs,v.(20)∫0∞∫0∞x3e−y2y2S2/πxk−log−x/y2logk−1−x/y2x7/2dxdy=1+i4k−1eiπk/2πk+1/2iζ−k,3/8−ζ−k,5/8−iζ−k,7/8+ζ−k,9/8.Proof.
Use equation (16), set m=−1,a=−1,b=1,α=1, and simplify using entry (3) in Table below (64 : 12 : 7) in [19].Example 5.
(21)∫0∞∫0∞x3e−y2y2S2/πxlog−x/y2+1x7/2log2−x/y2dxdy=4−4i−2π+2i2coth−128π.Proof.
Use equation (20), apply L’Hopital’s rule as k⟶−1, and simplify using equations (4.36.3) and (25.11.1) in [16].Example 6.
Bierens de Haan integral form.(22)∫0∞∫0∞1xe−y2S2/πxy−2m+py2mxpplogx/y2−1+xmy2p1−mlogx/y2x3log2x/y2dxdy=1π16+i6−eiπm/23iF2114,1;54;−e2imπ+eiπmF2134,1;74;−e2imπ+3ieiπp/2F2114,1;54;−e2ipπ+eiπp/2F2134,1;74;−e2ipπ.Proof.
Use equation (16) and form a second equation by replacing m⟶p and taking their difference. Next, set k=−1,a=1,b=1,α=1, and simplify using equation (9.559) in [14].Example 7.
(23)∫0∞∫0∞e−y2y4/3S2/πx8x−912y6logx/y2+12x12−y6x4x3log2x/y2dxdy=32πtanh−1−2+6−4sinπ/823−1sinπ/8−4.Proof.
Use equation (22), set m=−2/3,p=−3/4, and simplify.Example 8.
(24)∫0∞∫0∞e−y2y3/2S2/πx(16y−1510x20logx/y2+20y10−x20)20x3/10x3log2x/y2dxdy=−122πtanh−1222−2−8−210−2522−24−10−25−8.Proof.
Use equation (22), set m=−4/5,p=−3/4, and simplify.Example 9.
The Riemann zeta functionζs.(25)∫0∞∫0∞x3e−y2y2S2/πxk−logix/y2logk−1ix/y2x7/2dxdy=1+i2k−2eiπk/2πk+1/2i2kζ−k,14−i2kζ−k,34+2k+1−1ζ−k.Proof.
Use equation (16), set m=−1,a=i,b=α=1, and simplify using entry (4) in Table below (64 : 12 : 7) and entry (2) in Table (64 : 7) in [19].Example 10.
The constantlog2.(26)∫0∞∫0∞x3e−y2y2S2/πx1+logix/y2x7/2log2ix/y2dxdy=−1/16+i/16π+ilog4π.Proof.
Use equation (25), apply L’Hopital’s rule as k⟶−1, and simplify using equation (9.535.3) in [14].Example 11.
Apery’s constantζ3.(27)∫0∞∫0∞x3e−y2y2S2/πx3+logix/y2x7/2log4ix/y2dxdy=1/128+i/128π3+3iζ3π5/2.Proof.
Use equation (25), set k=−3, and simplify.Example 12.
Nested logarithmloglogx function.(28)∫0∞∫0∞x3e−y2y2S2/πx1−logix/y2loglogix/y2x7/2logix/y2dxdy=18π−iπ−1−ilogπΓ−1/429Γ−3/42+2+2ilog2.Proof.
Use equation (25) and take the first partial derivative with respect to k; set k=0 and simplify using (25.11.18) in [16].


## 6. Discussion

In this paper, we have presented a method for deriving a new double integral transform containing the Fresnel integralSx along with some interesting definite integrals, using contour integration. The results presented were numerically verified for both real and imaginary and complex values of the parameters in the integrals using Mathematica by Wolfram. We intend to pursue similar integrals containing the Fresnel function using our method.



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