# New Integrals Arising in the Samara-Valencia Heat Transfer Model in Grinding

Authors: J. L. González-Santander
Journal: Journal of Applied Mathematics
 (2017)

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## Abstract

The Samara-Valencia model for heat transfer in grinding has been recently used for calculating nontabulated integrals. Based on these results, new infinite integrals can be calculated, involving the Macdonald function and the modified Struve function.


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## Body

## 1. Introduction

Usually, mathematical developments facilitate the computation of mathematical modeling expressions in many different fields. However, mathematical modeling also yields in many cases a good field to develop new mathematical identities and formulas. This is the case of the mathematical modeling of heat transfer in surface grinding. This machining process consists in material removal from a workpiece by an abrasive wheel that rotates at high speed over its surface [1]. Classically, Jaeger’s model [2, 3] is used for the calculation of the temperature field in dry grinding. DesRuisseaux’s model [4] extends Jaeger’s model to include the effect of surface cooling (wet grinding). More recently, the Samara-Valencia model [5] has been proposed. In this model, the two-dimensional convective heat equation is considered. Also, the heat flux profile entering the workpiece and the action of the coolant are considered in the boundary condition. In [5], this boundary-value problem is transformed into an integral equation that is useful for the numerical evaluation of the heat transfer in intermittent wet grinding [6]. However, in the case of dry grinding, this integral equation can be reduced to a two-dimensional integral (T0theorem) [7].New mathematical identities have been proved in this framework. For instance, comparing Jaeger’s model with Samara-Valencia model, a new Dirac delta representation [8],(1)limu→0⁡uπK1x2+u2x2+u2=δx,and two new nontabulated integrals [9] have been obtained. Also, by using T0theorem, the following integrals have been calculated [10]:(2)∫0∞cosh⁡αξK0βξ2+y2dξ=πexp⁡-yβ2-α22β2-α2,β>α,y∈R,(3)K0x2+y2=yπ∫-∞∞K0ξ-xK1ξ2+y2ξ2+y2dξ,x,y∈R,(4)∫-∞∞K0ξK0x-ξdξ=π22e-x,x∈R.It is worth noting that (3) has been calculated also in [9] by using a complex integration contour. Recently, in [11], the following generalization of (4) has been calculated as a finite sum of terms containing beta and hypergeometric functions, by using the convolution theorem of the Fourier transform. This generalization reads as follows:(5)∫-∞∞t′α+2nKαat′t-t′β+2mKβat-t′dt′=πa2n!2m!2aμΓ2α+2n+1Γ2β+2m+1×∑k=0n+m-4kckn,m,α,β×12Bk+12,μ-k+12F12k+1212,k-μ+12a2t24+-1k+1at2μ-k+1sin⁡πα+βΓ2k-μ-1×F12μ+1μ-k+1,μ-k+32a2t24,Reα>-12-n,Reβ>-12-m,a>0,t∈R, where the following coefficients are defined as (6)ckn,m,α,β=∑l=max⁡k-m,0min⁡n,kbln,αbk-lm,β, with (7)bln,α=1n-l!2l!Γn+α-l+1.The scope of this paper is just to calculate more integrals based on results (2)–(4), which do not seem to be reported in the most common tables of integrals [12–14].This paper is organized as follows. Section2 is devoted to the calculation of the new integrals. It is divided into three subsections, each one of them containing one new result. Section 3 collects the conclusions, highlighting the main results obtained in the body of the paper.


## 2. The Integrals

### 2.1. First Result

Theorem 1.
The following integral holds true:(8)∫-∞∞e±αξK1βξ2+y2ξ2+y2dξ=2∫0∞cosh⁡αξK1βξ2+y2ξ2+y2dξ=πexp⁡-yβ2-α2βy,β>α,y∈R∖0.Proof.
Expanding the hyperbolic cosine ascosh⁡x=ex+e-x/2 [15, Eqn. 8.2] in (2), we have(9)πexp⁡-yβ2-α2β2-α2=∫0∞eαξK0βξ2+y2dξ+∫0∞e-αξK0βξ2+y2dξ.Now, perform the change of variables ξ→-ξ, and rewrite the first or the second integral given in (9) as(10)∫0∞e±αξK0βξ2+y2dξ=∫-∞0e∓αξK0βξ2+y2dξ.Therefore, inserting (10) in (9), we can define the following function:(11)Fy=∫-∞∞e±αξK0βξ2+y2dξ=πexp⁡-yβ2-α2β2-α2,β>α,y∈R. Let us take temporarily y>0, so that we can drop the absolute value in (11). Then, knowing that K0′x=-K1x [16, Eqn. 51:10:2], we have(12)F′y=βy∫-∞∞e±αξK1βξ2+y2ξ2+y2dξ=πexp⁡-yβ2-α2.Recover now the absolute value and take into account again (9)-(10) to obtain (8).Remark 2.
It is worth noting that wheny=0, (12) seems to fail. Nonetheless, this is apparent. In order to see it, rewrite the RHS of (12) as follows, knowing that β>0 and performing the changes of variables x=βx and u=βy:(13)F′u=u∫-∞∞exp⁡±αβxK1x2+u2x2+u2dx. Therefore, taking the limit u→0 and applying the Dirac delta representation given in (1), we have (14)limu→0⁡F′u=limu→0⁡u∫-∞∞exp⁡±αβxK1x2+u2x2+u2dx=π∫-∞∞exp⁡±αβxδxdx=π, which agrees with the LHS of (12), performing the limit y→0.We can continue calculating derivatives with respect toy in (12) in order to get new integrals, but the integrands we get are increasingly complex and we omit these results here.


### 2.2. Second Result

Theorem 3.
The following integral holds true:(15)∫0∞K0ξK0ξ2+y2dξ=π241-yK0yL-1y+K1yL0y,y∈R.Proof.
Let us define(16)Gy=∫0∞K0ξK0ξ2+y2dξ(17)=12∫-∞∞K0ξK0ξ2+y2dξ.Notice that(18)Gy=Gy;thereby, hereafter, we will assume that y>0, dropping the absolute value. Performing in (17) the derivative with respect to y, we have (19)G′y=-y2∫-∞∞K0ξK1ξ2+y2ξ2+y2dξ. The above integral can be calculated taking x=0 in (3); thus,(20)G′y=-π2K0y.Now, let us apply the following integral [12, Eqn. 1.12.1(3)]:(21)∫0zxνKνxdx=2ν-1πΓν+12zKνzLν-1z+Kν-1zLνz, where Lνz is the modified Struve function, defined as [17, Eqn. 11.2.2] (22)Lνz=∑n=0∞z/22n+ν+1Γn+3/2Γn+ν+3/2.Therefore, taking in (21) ν=0, knowing that K-νz=Kνz ([18], Eqn. 5.7.10) and also that [16, Eqn. 43:4:2](23)Γ12=π,we calculate Gy from (20) as(24)Gy=-π4yK0yL-1y+K1yLνy+C,where C is an integration constant. In order to calculate this integration constant, notice, on the one hand, that, from (16), we have (25)G0=∫0∞K02ξdξ. The above integral can be calculated taking x=0 in (4): (26)∫-∞∞K02ξdξ=2∫0∞K02ξdξ=π22. Thus,(27)G0=π24.On the other hand, taking limits in (24), we have(28)G0=C-π4limy→0⁡yK0yL-1y+K1yL0y.According to (22), consider the following asymptotic formula:(29)L-1y≈1Γ3/2Γ1/2=2π,y⟶0,where we have applied the property of the gamma function Γz+1=zΓz ([18], Eqn. 1.2.1) and (23). Similarly, we have(30)L0y≈y/2Γ3/2Γ1/2=yπ,y⟶0.Also, the asymptotic behavior of the Macdonald function for y→0 is ([17], Eqn. 10.30.2-3)(31)K0y≈-log⁡y,Kνy≈12Γνy2-ν,Re⁡ν>0, and thus(32)K1y≈1y.Therefore, taking into account (29)–(32), we calculate the limit given in (28) as(33)limy→0⁡yK0yL-1y+K1yL0y=1πlimy→0⁡y1-2log⁡y=0.Thereby, inserting (33) in (28) and recalling (27), we calculate the integration constant as follows:(34)G0=C=π24.Finally, substituting (34) into (24) and remembering the definition of Gy, we arrive at (15), where we have considered (18).


### 2.3. Third Result

Theorem 4.
The next integral holds true:(35)∫0∞1-yK0yL-1y+K1yL0ydy=2π.Proof.
For the third result, integrate both sides of (16) as follows:(36)∫0∞Gydy=∫0∞K0ξ∫0∞K0ξ2+y2dydξ.In order to calculate the inner integral given in (36), consider α=0 and β=1 in (2); hence,(37)∫0∞K0ξ2+y2dy=π2e-ξ.Inserting (37) into (36), we can drop the absolute value in the exponential, since the variable of integration is positive (i.e., ξ∈0,∞); thereby,(38)∫0∞Gydy=π2∫0∞K0ξe-ξdξ.To calculate the above integral, in the literature, we find the following integral, termedKing’s integral [19, Eqn. 11.3.16]: (39)∫0xe±uK0udu=xe±xK0x±K1x∓1. Hence, applying toKing’s Integral and the asymptotic formula [18, Eqn. 5.16.5] (40)Kνx≈π2xe-x,x⟶+∞, we obtain(41)∫0∞K0ξe-ξdξ=limx→∞⁡xe-xK0x-K1x+1=1.Recalling now the result given in (15) for Gy and taking into account (41), (38) finally reads as (35).


## 2.1. First Result

Theorem 1.
The following integral holds true:(8)∫-∞∞e±αξK1βξ2+y2ξ2+y2dξ=2∫0∞cosh⁡αξK1βξ2+y2ξ2+y2dξ=πexp⁡-yβ2-α2βy,β>α,y∈R∖0.Proof.
Expanding the hyperbolic cosine ascosh⁡x=ex+e-x/2 [15, Eqn. 8.2] in (2), we have(9)πexp⁡-yβ2-α2β2-α2=∫0∞eαξK0βξ2+y2dξ+∫0∞e-αξK0βξ2+y2dξ.Now, perform the change of variables ξ→-ξ, and rewrite the first or the second integral given in (9) as(10)∫0∞e±αξK0βξ2+y2dξ=∫-∞0e∓αξK0βξ2+y2dξ.Therefore, inserting (10) in (9), we can define the following function:(11)Fy=∫-∞∞e±αξK0βξ2+y2dξ=πexp⁡-yβ2-α2β2-α2,β>α,y∈R. Let us take temporarily y>0, so that we can drop the absolute value in (11). Then, knowing that K0′x=-K1x [16, Eqn. 51:10:2], we have(12)F′y=βy∫-∞∞e±αξK1βξ2+y2ξ2+y2dξ=πexp⁡-yβ2-α2.Recover now the absolute value and take into account again (9)-(10) to obtain (8).Remark 2.
It is worth noting that wheny=0, (12) seems to fail. Nonetheless, this is apparent. In order to see it, rewrite the RHS of (12) as follows, knowing that β>0 and performing the changes of variables x=βx and u=βy:(13)F′u=u∫-∞∞exp⁡±αβxK1x2+u2x2+u2dx. Therefore, taking the limit u→0 and applying the Dirac delta representation given in (1), we have (14)limu→0⁡F′u=limu→0⁡u∫-∞∞exp⁡±αβxK1x2+u2x2+u2dx=π∫-∞∞exp⁡±αβxδxdx=π, which agrees with the LHS of (12), performing the limit y→0.We can continue calculating derivatives with respect toy in (12) in order to get new integrals, but the integrands we get are increasingly complex and we omit these results here.


## 2.2. Second Result

Theorem 3.
The following integral holds true:(15)∫0∞K0ξK0ξ2+y2dξ=π241-yK0yL-1y+K1yL0y,y∈R.Proof.
Let us define(16)Gy=∫0∞K0ξK0ξ2+y2dξ(17)=12∫-∞∞K0ξK0ξ2+y2dξ.Notice that(18)Gy=Gy;thereby, hereafter, we will assume that y>0, dropping the absolute value. Performing in (17) the derivative with respect to y, we have (19)G′y=-y2∫-∞∞K0ξK1ξ2+y2ξ2+y2dξ. The above integral can be calculated taking x=0 in (3); thus,(20)G′y=-π2K0y.Now, let us apply the following integral [12, Eqn. 1.12.1(3)]:(21)∫0zxνKνxdx=2ν-1πΓν+12zKνzLν-1z+Kν-1zLνz, where Lνz is the modified Struve function, defined as [17, Eqn. 11.2.2] (22)Lνz=∑n=0∞z/22n+ν+1Γn+3/2Γn+ν+3/2.Therefore, taking in (21) ν=0, knowing that K-νz=Kνz ([18], Eqn. 5.7.10) and also that [16, Eqn. 43:4:2](23)Γ12=π,we calculate Gy from (20) as(24)Gy=-π4yK0yL-1y+K1yLνy+C,where C is an integration constant. In order to calculate this integration constant, notice, on the one hand, that, from (16), we have (25)G0=∫0∞K02ξdξ. The above integral can be calculated taking x=0 in (4): (26)∫-∞∞K02ξdξ=2∫0∞K02ξdξ=π22. Thus,(27)G0=π24.On the other hand, taking limits in (24), we have(28)G0=C-π4limy→0⁡yK0yL-1y+K1yL0y.According to (22), consider the following asymptotic formula:(29)L-1y≈1Γ3/2Γ1/2=2π,y⟶0,where we have applied the property of the gamma function Γz+1=zΓz ([18], Eqn. 1.2.1) and (23). Similarly, we have(30)L0y≈y/2Γ3/2Γ1/2=yπ,y⟶0.Also, the asymptotic behavior of the Macdonald function for y→0 is ([17], Eqn. 10.30.2-3)(31)K0y≈-log⁡y,Kνy≈12Γνy2-ν,Re⁡ν>0, and thus(32)K1y≈1y.Therefore, taking into account (29)–(32), we calculate the limit given in (28) as(33)limy→0⁡yK0yL-1y+K1yL0y=1πlimy→0⁡y1-2log⁡y=0.Thereby, inserting (33) in (28) and recalling (27), we calculate the integration constant as follows:(34)G0=C=π24.Finally, substituting (34) into (24) and remembering the definition of Gy, we arrive at (15), where we have considered (18).


## 2.3. Third Result

Theorem 4.
The next integral holds true:(35)∫0∞1-yK0yL-1y+K1yL0ydy=2π.Proof.
For the third result, integrate both sides of (16) as follows:(36)∫0∞Gydy=∫0∞K0ξ∫0∞K0ξ2+y2dydξ.In order to calculate the inner integral given in (36), consider α=0 and β=1 in (2); hence,(37)∫0∞K0ξ2+y2dy=π2e-ξ.Inserting (37) into (36), we can drop the absolute value in the exponential, since the variable of integration is positive (i.e., ξ∈0,∞); thereby,(38)∫0∞Gydy=π2∫0∞K0ξe-ξdξ.To calculate the above integral, in the literature, we find the following integral, termedKing’s integral [19, Eqn. 11.3.16]: (39)∫0xe±uK0udu=xe±xK0x±K1x∓1. Hence, applying toKing’s Integral and the asymptotic formula [18, Eqn. 5.16.5] (40)Kνx≈π2xe-x,x⟶+∞, we obtain(41)∫0∞K0ξe-ξdξ=limx→∞⁡xe-xK0x-K1x+1=1.Recalling now the result given in (15) for Gy and taking into account (41), (38) finally reads as (35).


## 3. Conclusions

Based on integrals (2)–(4), calculated in the framework of the Samara-Valencia heat transfer model in surface grinding by using the T0theorem, new integrals have been derived. From (2), we have derived integrals (11) and (8). Also, applying (3) and (4), we have derived (15). Finally, integrating the result given in (15) and taking into account (2), the integral given in (35) has been obtained. It is worth noting that these results have been confirmed, evaluating numerically the corresponding integrals.



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