Brian Fisher

Fisher Brian

Joel D. Nicholas

Nicholas Joel D.

Fisher, Brian

Nicholas, Joel D.

Tokyo University of Science Repository for Academic Resources / 東京理科大学学術リポジトリ

SUT Journal of MathematicsVol. 33, No. 2 (1997), 139{148THE EXPONENTIAL INTEGRAL AND THECONVOLUTIONBrian Fisher and Joel D. Nicholas(Received June 16, 1997)Abstract. The exponential integral ei(¸x) and its associated functions ei+(¸x)and ei¡(¸x) are de¯ned as locally summable functions on the real line andtheir derivatives are found as distributions. Some convolution products of thesedistributions and other distributions are then found.AMS 1991 Mathematics Subject Classi¯cation. 33B10, 46F10.Key words and phrases. exponential integral, convolution product.The exponential integral ei(x) is de¯ned for x > 0 byei(x) =Z 1xu¡1e¡udu; (1)see Sneddon [3], the integral diverging for x · 0: It was pointed out in [1]that equation (1) can be rewritten in the formei(x) =Z 1xu¡1[e¡u ¡H(1¡ u)] du¡H(1¡ x) ln jxj;where H denotes Heaviside's function. The integral in this equation is con-vergent for all x and so was used to de¯ne ei(x) on the real line.More generally, if ¸ 6= 0, ei(¸x) was de¯ned in the obvious way byei(¸x) =Z 1¸xu¡1[e¡u ¡H(1¡ u)] du¡H(1¡ ¸x) ln j¸xj: (2)Further, ei+(¸x) and ei¡(¸x) were de¯ned byei+(¸x) = H(x) ei(¸x); ei¡(¸x) = H(¡x) ei(¸x)139140 B. FISHER AND J. D. NICHOLASso thatei(¸x) = ei+(¸x) + ei¡(¸x): (3)In particular, if ¸ > 0, we haveei(¸x) =Z 1xu¡1[e¡¸u ¡H(1¡ ¸u)] du¡H(1¡ ¸x) ln j¸xj; (4)ei+(¸x) =Z 1xu¡1e¡¸u du; x > 0; (5)ei¡(¸x) = ¡° ¡ ln j¸j+Z 0xu¡1(e¡¸u ¡ 1) du¡ lnx¡; x < 0; (6)where° = ¡Z 10u¡1[e¡¸u ¡H(1¡ ¸u)] duis Euler's constant.If ¸ < 0, we haveei(¸x) = ¡Z x¡1u¡1[e¡¸u ¡H(1¡ ¸u)] du¡H(1¡ ¸x) ln j¸xj; (7)ei+(¸x) = ¡° ¡ ln j¸j ¡Z x0u¡1(e¡¸u ¡ 1) du¡ lnx+; x > 0; (8)ei¡(¸x) = ¡Z x¡1u¡1e¡¸u du; x < 0: (9)The derivatives of these functions were found as[ei(¸x)]0 = ¡e¡¸xx¡1 = ¡x¡1 ¡1Xi=1(¡¸)ii!xi¡1; (10)[ei+(¸x)]0 = ¡e¡¸xx¡1+ ¡ (° + ln j¸j)±(x)= ¡x¡1+ ¡1Xi=1(¡¸)ii!xi¡1+ ¡ (° + ln j¸j)±(x); (11)[ei¡(¸x)]0 = e¡¸xx¡1¡ + (° + ln j¸j)±(x)= x¡1¡ ¡1Xi=1¸ii!xi¡1¡ + (° + ln j¸j)±(x); (12)for all ¸ 6= 0:We now note the following results obtained by replacing x by ¡x in thefunctions ei(¸x), ei+(¸x) and ei¡(¸x).ei(¸(¡x)) = ei((¡¸)x); (13)ei+(¸(¡x)) = H(¡x) ei(¸(¡x)) = ei¡((¡¸)x); (14)ei¡(¸(¡x)) = H(x) ei(¸(¡x)) = ei+((¡¸)x): (15)THE EXPONENTIAL INTEGRAL AND THE CONVOLUTION 141These results will be used to deduce results for ¸ < 0 from results proved for¸ > 0:The classical de¯nition of the convolution product of two functions f andg is as follows:De¯nition 1. Let f and g be functions. Then the convolution product f ¤ gis de¯ned by(f ¤ g)(x) =Z 1¡1f(t)g(x¡ t) dtfor all points x for which the integral exist.It follows easily from the de¯nition that if f ¤ g exists then g ¤ f exists andf ¤ g = g ¤ f (16)and if (f ¤ g)0 and f ¤ g0 (or f 0 ¤ g) exists, then(f ¤ g)0 = f ¤ g0 (or f 0 ¤ g): (17)De¯nition 1 can be extended to de¯ne the convolution product f ¤ g oftwo distributions f and g in D0 with the following de¯nition, see Gel'fand andShilov [2].De¯nition 2. Let f and g be distributions in D0. Then the convolutionproduct f ¤ g is de¯ned by the equationh(f ¤ g)(x); Ái = hf(y); hg(x); Á(x+ y)iifor arbitrary Á in D, provided f and g satisfy either of the conditions(a) either f or g has bounded support,(b) the supports of f and g are bounded on the same side.It follows that if the convolution product f ¤g exists by this de¯nition thenequations (16) and (17) are satis¯ed. In the following, the locally summablefunctions e¸x+ and e¸x¡ are de¯ned for ¸ 6= 0 bye¸x+ = H(x)e¸x e¸x¡ = H(¡x)e¸x:Note thate¸(¡x) = e(¡¸)x; e¸(¡x)+ = e(¡¸)x¡ ; e¸(¡x)¡ = e(¡¸)x+ : (18)These results will also be used to deduce results for ¸ < 0 from results provedfor ¸ > 0:We now prove the following theorem142 B. FISHER AND J. D. NICHOLASTheorem 1. If ¸ 6= 0 and ¹ 6= 0, then the convolution product ei+(¸x)¤e¹x+exists andei+(¸x) ¤ e¹x+ = ¹¡1fe¹x ei+[(¸+ ¹)x] + ln j1 + ¹=¸je¹x+ ¡ ei+(¸x)g (19)if ¸+ ¹ 6= 0 andei+(¸x) ¤ e¡¸x+ = ¸¡1[ei+(¸x) + (° + ln j¸j)e¡¸x+ + e¡¸x lnx+]: (20)if ¸+ ¹ = 0.Proof. The convolution product ei+(¸x) ¤ e¹x+ = 0 if x < 0 and so we supposethat x > 0: There are four cases to consider to prove equation (19).Case (i). ¸ > 0; ¸+ ¹ > 0:We ¯rst of all prove thatei+(¸x) ¤ e¹x+ = ¹¡1e¹x+Z x0u¡1[e¡¸u ¡ e¡(¸+¹)u] du++¹¡1(e¹x ¡ 1) ei+(¸x): (21)We haveei+(¸x) ¤ e¹x+ =Z x0e¹(x¡t)Z 1tu¡1e¡¸u du dt=Z x0u¡1e¡¸uZ u0e¹(x¡t) dt du+Z 1xu¡1e¡¸uZ x0e¹(x¡t) dt du= ¹¡1e¹x+Z x0u¡1[e¡¸u ¡ e¡(¸+¹)u] du++¹¡1(e¹x ¡ 1) ei+(¸x);giving equation (21).Further,Z x0u¡1[e¡¸u ¡H(1¡ ¸u)] du =Z 10u¡1[e¡¸u ¡H(1¡ ¸u)] du+¡Z 1xu¡1e¡¸u du+Z 1xu¡1H(1¡ ¸u) du= ¡° ¡ ei+(¸x) +Z 1xu¡1H(1¡ ¸u) du:(22)SimilarlyZ x0u¡1[e¡(¸+¹)u ¡H(1¡ (¸+ ¹)u)] du = ¡° ¡ ei+[(¸+ ¹)x] ++Z 1xu¡1H[1¡ (¸+ ¹)u] du:(23)THE EXPONENTIAL INTEGRAL AND THE CONVOLUTION 143It follows from equations (22) and (23) thatZ x0u¡1(e¡¸u ¡ e¡(¸+¹)u) du = ei+[(¸+ ¹)x]¡ ei+(¸x) ++Z 10u¡1[H(1¡ ¸u)¡H(1¡ (¸+ ¹)u)] du= ei+[(¸+ ¹)x]¡ ei+(¸x) + ln(1 + ¹=¸): (24)Equation (19) now follows from equations (21) and (24) for Case (i).Case (ii). ¸ > 0; ¸+ ¹ < 0:Equations (21) and (22) again hold in this case but since ¸ + ¹ < 0; wehave from equation (8)Z x0u¡1(e¡(¸+¹)u ¡ 1) du = ¡° ¡ ln j¸+ ¹j ¡ ei+[(¸+ ¹)x]¡ lnx+: (25)It follows from equations (22) and (25) that equation (24) again holds. Equa-tion (19) follows for Case (ii).Case (iii). ¸ < 0; ¸+ ¹ < 0:This time we haveei+(¸x) ¤ e¹x+ = ¡(° + ln j¸j)Z x0e¹tdt¡Z x0e¹(x¡t)Z t0u¡1(e¡¸u ¡ 1) du dt+¡Z x0e¹(x¡u) lnu du= ¡¹¡1(° + ln j¸j)(e¹x ¡ 1)¡Z x0u¡1(e¡¸u ¡ 1)Z xue¹(x¡t) dt du++¹¡1e¹xZ x0lnu d(e¡¹u ¡ 1)= ¡¹¡1(° + ln j¸j)(e¹x ¡ 1) ++¹¡1Z x0u¡1(e¡¸u ¡1)(1¡ e¹(x¡u)) du++¹¡1(1¡ e¹x) lnx¡ ¹¡1e¹xZ x0u¡1(e¡¹u ¡ 1) du= ¡¹¡1 ei+(¸x)¡ ¹¡1e¹xZ x0u¡1(e¡(¸+¹)u ¡ e¡¹u) du+¡¹¡1(° + ln j¸j)e¹x ¡ ¹¡1e¹x lnx¡ ¹¡1e¹xZ x0u¡1(e¡¹u ¡ 1) du= ¡¹¡1 ei+(¸x)¡ ¹¡1e¹xZ x0u¡1(e¡(¸+¹)u ¡ 1) du+¡¹¡1(° + ln j¸j)e¹x ¡ ¹¡1e¹x lnx (26)and equation (19) follows for Case (iii).144 B. FISHER AND J. D. NICHOLASCase (iv). ¸ < 0; ¸+ ¹ > 0:Equation (26) still holds for this case but this time we haveZ x0u¡1(e¡(¸+¹)u ¡ 1) du =Z 10u¡1[e¡(¸+¹)u ¡H(1¡ (¸+ ¹)u)] du+¡Z 1xu¡1e¡(¸+¹)u du+Z (¸+¹)¡1xu¡1 du= ¡° ¡ ei+[(¸+ ¹)x]¡ ln[(¸+ ¹)x]and equation (19) now follows from this equation and equation (26) for Case(iv).We now have a further two cases to consider when ¸+ ¹ = 0.Case (v). ¸ > 0; ¸+ ¹ = 0:Equation (21) holds for this case. Further, replacing ¸+¹ by ¹ in equation(25) we have Z x0u¡1(e¡¸u ¡ 1) du = ¡° ¡ ei+(¸x)¡ ln(¸x) (27)and equation (20) now follows from equation (21) for Case (v).Case (vi). ¸ < 0; ¸+ ¹ = 0:Equation (26) holds when ¹ = ¡¸ but it reduces toei+(¸x) ¤ e¡¸x+ = ¸¡1 ei+(¸x) + ¸¡1(° + ln j¸j)e¡¸x + ¸¡1e¡¸x lnxand equation (20) follows for Case (vi). 2Corollary 1.1. If ¸ 6= 0 and ¹ 6= 0, then the convolution product (e¡¸xx¡s+ )¤e¹x+ exists for s = 1; 2; : : : : In particular, if ¸+ ¹ 6= 0; then(e¡¸xx¡1+ ) ¤ e¹x+ = ¡e¹x ei+[(¸+ ¹)x]¡ (° + ln j¸+ ¹j)e¹x+ (28)and if ¸+ ¹ = 0, then(e¡¸xx¡1+ ) ¤ e¡¸x+ = e¡¸x lnx+: (29)Proof. The convolution product (e¡¸xx¡s+ ) ¤ e¹x+ exists by De¯nition 2 fors = 1; 2; : : : since e¡¸xx¡s+ and e¹x+ are both bounded on the left. In particular,we have from equations (11), (17) and (19)[¡e¡¸xx¡1+ ¡ (° + ln j¸j)±(x)] ¤ e¹x+ = ei+(¸x) ¤ [¹e¹x+ + ±(x)]= e¹x ei+[(¸+ ¹)x] + ln j1 + ¹=¸je¹x+THE EXPONENTIAL INTEGRAL AND THE CONVOLUTION 145and equation (28) follows.Similarly, using equations (11), (17) and (20), we have[¡e¡¸xx¡1+ ¡ (° + ln j¸j)±(x)] ¤ e¡¸x+ = ei+(¸x) ¤ [¡¸e¡¸x+ + ±(x)]= ¡ ei+(¸x)¡ (° + ln j¸j)e¡¸x+ ¡ e¡¸x lnx+ + ei+(¸x)and equation (29) follows. 2Theorem 2. If ¸ 6= 0 and ¹ 6= 0, then the convolution product ei¡(¸x)¤e¹x¡exists andei¡(¸x) ¤ e¹x¡ = ¡¹¡1fe¹x ei¡[(¸+ ¹)x] + ln j1 + ¹=¸je¹x¡ ¡ ei¡(¸x)g (30)if ¸+ ¹ 6= 0, andei¡(¸x) ¤ e¡¸x¡ = ¡¸¡1[ei¡(¸x) + (° + ln j¸j)e¡¸x¡ + e¡¸x lnx¡] (31)if ¸+ ¹ = 0:Proof. Replacing ¸ by ¡¸ and ¹ by ¡¹ in equation (19) we getei+((¡¸)x) ¤ e(¡¹)x+ = ¡¹¡1fe(¡¹)x ei+[((¡¸¡ ¹)x] + ln j1 + ¹=¸)je(¡¹)x+ +¡ ei+((¡¸)x)gand equation (30) follows on replacing x by ¡x in this equation.Equation (31) follows similarly. 2Corollary 2.1. If ¸ 6= 0 and ¹ 6= 0, then the convolution product (e¡¸xx¡s¡ )¤e¹x¡ exists for s = 1; 2; : : : : In particular, if ¸+ ¹ 6= 0; then(e¡¸xx¡1¡ ) ¤ e¹x¡ = ¡e¹x ei¡[(¸+ ¹)x]¡ (° + ln j¸+ ¹j)e¹x¡ (32)and if ¸+ ¹ = 0 then(e¡¸xx¡1¡ ) ¤ e¡¸x¡ = e¡¸x lnx¡: (33)Proof. The existence of convolution product (e¡¸xx¡s¡ ) ¤ e¹x¡ follows fromequations (11), (17) and (30). In particular, we have from equations (11), (17)and (30)[e¡¸xx¡1¡ + (° + ln j¸j)±(x)] ¤ e¹x¡ = ei¡(¸x) ¤ [¹e¹x¡ ¡ ±(x)]= ¡e¹x ei¡[(¸+ ¹)x]¡ ln(1 + ¹=¸)e¹x¡and equation (32) follows. Similarly, using equations (11), (17) and (31), wehave[e¡¸xx¡1¡ + (° + ln j¸j)±(x)] ¤ e¡¸x¡ = ei¡(¸x) ¤ [¡¸e¸x¡ ¡ ±(x)]= (° + ln j¸j)e¡¸x¡ + e¡¸x lnx¡and equation (33) follows. 2146 B. FISHER AND J. D. NICHOLASTheorem 3. If ¸; ¸ + ¹ > 0 and ¹ 6= 0, then the convolution productei+(¸x) ¤ e¹x exists andei+(¸x) ¤ e¹x = ¹¡1 ln(1 + ¹=¸)e¹x: (34)Proof. We haveei+(¸x) ¤ e¹x =Z 10e¹(x¡t)Z 1tu¡1e¡¸u du dt=Z 10u¡1e¡¸uZ u0e¹(x¡t) dt du= ¹¡1e¹xZ 10u¡1[e¡¸u ¡ e¡(¸+¹)u] du:NowZ 10u¡1[e¡¸u ¡ e¡(¸+¹)u] du =Z 10u¡1[e¡¸u ¡H(1¡ ¸u)] du+¡Z 10u¡1[e¡(¸+¹)u ¡H(1¡ (¸+ ¹)u)] du++Z 10u¡1[H(1¡ ¸u)¡H(1¡ (¸+ ¹)u)] du= ¡° + ° + ln(1 + ¹=¸)and equation (34) follows. 2Note 1. Theorem 3 is equivalent to the van der Pol formula [4]Z 10e¡px ei(¸x) dx = p¡1 ln(1 + p=¸):Corollary 3.1. If ¸; ¸ + ¹ > 0 and ¹ 6= 0, then the convolution product(e¡¸xx¡1+ ) ¤ e¹x exists and(e¡¸xx¡1+ ) ¤ e¹x = ¡(° + ln j¸+ ¹j)e¹x: (35)Proof. Di®erentiating equation (34) we get[¡e¡¸xx¡1+ ¡ (° + ln j¸j)±(x)] ¤ e¹x = ln(1 + ¹=¸)e¹xand equation (35) follows. 2Note 2. Corollary 3.1 is equivalent toZ 1¡1e¡pxx¡1+ dx = ¡° ¡ ln p;due to Gel'fand and Shilov [2].THE EXPONENTIAL INTEGRAL AND THE CONVOLUTION 147Corollary 3.2. If ¸; ¸ + ¹ > 0 and ¹ 6= 0, then the convolution productsei+(¸x) ¤ e¹x¡ and (e¡¸xx¡1+ ) ¤ e¹x¡ exist andei+(¸x) ¤ e¹x¡ = ¹¡1fei+(¸x)¡ e¹x ei+[(¸+ ¹)x] + ln(1 + ¹=¸)e¹x¡ g(36)(e¡¸xx¡1+ ) ¤ e¹x¡ = e¹x ei+[(¸+ ¹)x]¡ (° + ln j¸+ ¹j])e¹x¡ : (37)Proof. Equation (36) follows from equations (19) and (34). Equation (37)then follows from equations (28) and (35). 2Theorem 4. If ¸; ¸ + ¹ < 0 and ¹ 6= 0, then the convolution productei¡(¸x) ¤ e¹x exists andei¡(¸x) ¤ e¹x = ¡¹¡1 ln(1 + ¹=¸)e¹x: (38)Proof. Replacing ¸ by ¡¸ and ¹ by ¡¹ in equation (34) we getei+[(¡¸)x)] ¤ e¡¹x = ¡¹¡1 ln(1 + ¹=¸)e¡¹xand equation (38) follows on replacing x by ¡x in this equation. 2The results of the corollaries follow easily.Corollary 4.1. If ¸; ¸ + ¹ < 0 and ¹ 6= 0, then the convolution product(e¡¸xx¡1¡ ) ¤ e¹x exists and(e¡¸xx¡1¡ ) ¤ e¹x = ¡(° + ln j¸+ ¹j)e¹x:Corollary 4.2. If ¸; ¸ + ¹ < 0 and ¹ 6= 0, then the convolution productsei¡(¸x) ¤ e¹x+ and (e¡¸xx¡1¡ ) ¤ e¹x+ exist andei¡(¸x) ¤ e¹x+ = ¹¡1fei¡(¸x)¡ e¹x ei¡[(¸+ ¹)x] + ln(1 + ¹=¸)e¹x+ g(e¡¸xx¡1¡ ) ¤ e¹x+ = e¹x ei¡[(¸+ ¹)x]¡ (° + ln j¸+ ¹j)e¹x+ :Acknowledgment. The authors would like to thank the referee for his helpfulsuggestions in the improvement of this paper.References[1] B. Fisher and J.D Nicholas, \On the exponential integral", submitted for publi-cation.[2] I.M. Gel'fand and G.E. Shilov, \Generalized functions", Vol. I, Academic Press(1964).148 B. FISHER AND J. D. NICHOLAS[3] I.N. Sneddon, \Special Functions of Mathematical Physics and Chemistry",Oliver and Boyd, (1961).[4] B. van der Pol, \On the operational solution of linear di®erential equations andan investigation of the properties of these solutions", Phil. Mag. ser.7 8(1929),861-898.Brian Fisher and Joel D. NicholasDepartment of Mathematics and Computer Science, Leicester UniversityLeicester, LE1 7RH, England.

THE EXPONENTIAL INTEGRAL AND THE CONVOLUTION

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THE EXPONENTIAL INTEGRAL AND THE CONVOLUTION

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