The integrals in Gradshteyn and Ryzhik. Part 10: The digamma function

Transcription

1 SCIENTIA Series A: Mathematical Sciences, Vol 7 29, Universidad Técnica Federico Santa María Valparaíso, Chile ISSN c Universidad Técnica Federico Santa María 29 The integrals in Gradshteyn and Ryzhi Part : The digamma function Luis A Medina and Victor H Moll Abstract The table of Gradshteyn and Rhyzi contains some integrals that can be expressed in terms of the digamma function ψx = d log Γx In this note dx we present some of these evaluations Introduction The table of integrals [2] contains a large variety of definite integrals that involve the digamma function ψx = d dx log Γx = Γ x Γx Here Γx is the gamma function defined by 2 Γx = t x e t dt Many of the analytic properties can be derived from those of Γx The next theorem represents a collection of the important properties of Γx that are used in the current paper The reader will find in [] detailed proofs Theorem The gamma function satisfies: a the functional equation 3 Γx + = xγx b For n N, the interpolation formula Γn = n! 2 Mathematics Subject Classification Primary 33 Key words and phrases Integrals, Digamma function The second author wishes to acnowledge the partial support of NSF-DMS The first author was partially supported as a graduate student by the same grant 9

2 L MEDINA AND V MOLL c The Euler constant γ, defined by 4 γ = lim n n = lnn, is also given by γ = Γ This appears as the special case a = of formula 433: 5 e ax = γ + lna a This was established in [3] The change of variables t = ax shows that the case a = is equivalent to the general case This is an instance of a fae parameter d The infinite product representation 6 Γx = e γx x = [ + x e x/] is valid for x C away from the poles at x =,, 2, e For n N we have 7 Γ n + 2 and 2n! = π 2 2n n! 8 Γ 2 n = n 22n n! π 2n! f For x C, x Z we have the reflection rule 9 ΓxΓ x = π sin πx Several properties of the digamma function ψx follow directly from the gamma function Theorem 2 The digamma function ψx satisfies a the functional equation ψx + = ψx + x b For n N, we have n ψn = γ + In particular, ψ = γ =

3 DIGAMMA FUNCTION c For x C away from x =,, 2, we have 2 d The derivative of ψ is given by 3 ψ x = In particular, ψ = π 2 /6 e For n N we have ψx = γ x + x x + = = γ x + + = = x ψ 2 ± n = γ 2 ln n In particular, 5 ψ 2 = γ 2 ln2 f For x C, x Z we have the reflection rule = 6 ψ x = ψx + π cot πx 2 2 A first integral representation In this section we establish the integral evaluation 3429 Several direct consequences of this formulas are also described Proposition 2 Assume a > Then [ 2 e x + x a] dx x = ψa 22 Proof We begin with the double integral On the other hand, 23 We conclude that 24 s s e tz dt dz = This evaluation is equivalent to: 25 e tz dz dt = s e z e sz dz z dt t = lns e z e sz dz = lns z e ax e bx dx = ln b x a,

4 2 L MEDINA AND V MOLL that appears as formula in [2] The reader will find a proof in [3] We now establish the result: start with Γ a = = = This formula can be rewritten as This establishes 2 e s s a lnsds e s s a e z e zs dz ds z dz e z s a e s ds s a e s+z ds z Γ a = Γa e z + z a dz z Example 22 The special case a = yields 26 e x dx + x x = γ This appears as Example 23 The change of variables w = lnx gives the value of 42752: [ q ] dx 27 x lnx x lnx = [ e w + w q] dw w = ψq 28 Example 24 The change of variables t = /x + in 2 yields 3474: e /t t a t t dt = ψa Example 25 The result of Example 22 can be used to prove 34354: 29 e bx dx + ax x = ln a b γ Indeed, the change of variables t = bx yields from 22 the identity e bx dx + ax x = = e t dt + at/b t e t e at/b dt + t e at/b dt + at/b t Formula 25 shows the first integral is ln a b and the value of the second one comes from 22

5 DIGAMMA FUNCTION 3 Example 26 The evaluation 34762: 2 e xp e xq dx x = p q pq γ comes directly from 2 Indeed, the change of variables u = x p yields I := e xp e xq dx x = e u e uq/p du p u Now write I = p e u du + u u + p + u e uq/p du u The first integral is γ by 26 and the change of variables v = u q/p gives I = γ p + dv e v q + vp/q v = γ p + dv q + v e v v + v v p/q q v + v + v p/q dv Split the last integral from [, ] to [, and use the change of variables x /x in the second part to chec that the whole integral vanishes Formula 2 has been established Example 27 Formula 3463: 2 e x2 e x dx x = γ 2 corresponds to the choice p = 2 and q = in 2 Example 28 Formula 34692: 22 e x4 e x dx x = 3γ 4 corresponds to the choice p = 4 and q = in 2 Example 29 Formula 34693: 23 e x4 e x2 dx x = γ 4 corresponds to the choice p = 4 and q = 2 in 2 Example 2 Formula 34753: 24 e x2n e x dx x = 2 n γ corresponds to the choice p = 2 n and q = in 2

6 4 L MEDINA AND V MOLL The case p = q in 2 is now modified to include a parameter Proposition 2 Let a, b, p R + Then 3476 in [2] states that [ 25 e axp e bxp] dx ln b lna = x p Proof The change of variables t = ax p gives [ e axp e bxp] dx x = e t e bt/a dt p t Introduce the term / + t to obtain I = e t dt p + t t p e s = γ p p b b + as ds s Adding and subtracting the term / + s produces 26 I = b p b + as ds + s s The final result now comes from evaluating the last integral e bt/a dt + t t We now present another integral representation of the digamma function Proposition 22 The digamma function is given by e x 27 ψa = x e ax e x dx This expression appears as 3427 in [2] Proof The representation 2 is written as 28 ψa = lim δ e z δ z dz δ dz z + z a, to avoid the singularity at z = The change of variables z = e t in the second integral gives 29 ψa = lim δ Now observe that 22 e z δ ln+δ e t as δ This completes the proof δ t z dz dt δ ln+δ ln+δ e at dt e t dt t,

7 DIGAMMA FUNCTION 5 Example 23 The special case a = in 27 gives 34272: 22 e x e x dx = γ x Example 24 The change of variables t = e x in 27 produces 4284: 222 lnt + ta dt = ψa t Example 25 The special case a = in 222 yields 428: 223 lnt + dt = γ t Proposition 26 Let p, q R Then x p 224 lnx + xq dx = lnp ψq x This appears as 4285 in [2] Proof Write x p 225 lnx + xq dx = x lnx + xq x p dx + dx x lnx The first integral is ψq from 222 and to evaluate the second one, differentiate with respect to p, to produce 226 d dp x p lnx dx = x p dx = p The value at p = shows that the constant of integration vanishes The formula 224 has been established 3 The difference of values of the digamma function In this section we establish an integral representation for the difference of values of the digamma function The expression appears as 3235 in [2] 3 Proposition 3 Let p, q R Then Proof Consider first 32 Iǫ = x p x q dx = ψq ψp x x p x ǫ dx x q x ǫ dx, that avoids the apparent singularity at x = The integral Iǫ can be expressed in terms of the beta function 33 Ba, b = x a x b dx

8 6 L MEDINA AND V MOLL as Iǫ = Bp, ǫ Bq, ǫ, and using the relation 34 Ba, b = ΓaΓb Γa + b we obtain 35 Iǫ = Γǫ Now use Γ + ǫ = ǫγǫ to write Γp Γp + ǫ 36 Iǫ = Γ + ǫ ǫ and obtain 3 by letting ǫ Γp Γp + ǫ Γq Γq + ǫ Γq Γq + ǫ Γp + ǫ ǫ, Γq + ǫ 37 Example 32 The special value ψ = γ produces This appears as 3265 in [2] 38 x q x dx = γ + ψq Example 33 A second special value appears in 32682: x a x xb dx = ψa + b ψb It is obtained from 3 by choosing p = b and q = a + b 39 Example 34 Now let q = p in 3 to produce This appears as 323 in [2] 3 x p x p dx = ψ p ψp = π cot πp x Example 35 The special case p = a + and q = a produces x a x a x dx = ψ a ψ + a = π cot πa a, where we have used and 6 to simplify the result This is 3233 in [2] 3 Example 36 The change of variables x = t a in 3 produces t ap t aq t a dt = ψq ψp a Now let p =, a = ν and q = µ ν and the replace µ by p and ν by q to obtain in [2]: t q t p 32 t q dt = p γ + ψ q q

9 DIGAMMA FUNCTION 7 33 Example 37 The special case p = b/a and q = b/a in 3 produces x b x a b x a dx = ψ b/a ψb/a a The result is now simplified using 6 to produce 34 This is in [2] 35 x b x a b x a dx = π a cot πb a Example 38 The special case a = 2 in 3 yields The choice µ = + p/2 and ν = p/2: 36 t 2µ t 2ν t 2 dt = ψν ψµ 2 x p x p x 2 xdx = ψ + p/2 ψ p/2 2 The identities ψx + = ψx + /x and ψ x ψx = π cotπx produce x p x p 37 x 2 xdx = π pπ 2 cot 2 p This appears as 3269 in [2] Example 39 The choice µ = a+ 2 and ν = b+ 2 in 35 gives 32693: x a x b 38 x 2 dx = b + a + ψ ψ Integrals over a half-line In this section we consider integrals over the half-line [, that can be evaluated in terms of the digamma function Proposition 4 Let p, q R Then t p 4 + t p t q dt + t q t This is 329 in [2] Also 42 + t p + t q dt t = ψq ψp = ψq ψp Proof Let t = x/ x in 3 The second form comes from the first by the change of variables x /x Example 42 The special case p = yields 43 + t dt + t q t This appears as 3233 in [2] = ψq + γ

10 8 L MEDINA AND V MOLL 44 Example 43 The evaluation of 3235: + x a dx + x b = ψb ψb a x can be established directly from 43 Simply write + x a dx + x b x = + x dx + x b x to obtain the result + x dx + x b a x Some examples of integrals over [, can be reduced to a pair of integrals over [, ] 45 Proposition 44 The formula 3236 of [2] states that x p x q dx = π cotπp cotπq x Proof To evaluate this, mae the change of variables t = /x in the part over [, to produce x p x q x p x q x p x q 46 dx = dx dx x x x Now use the result 3 to write 47 x p x q dx = ψq ψp [ψ q ψ p] x The relation ψx ψ x = π cotπq yields the result 5 An exponential scale In this section we present the evaluation of certain definite integrals involving the exponential function These are integrals that can be evaluated in terms of the digamma function of the parameters involved Example 5 The simplest one is 3372: 5 + e x p + e x q dx = ψq ψp that comes from 42 via the change of variables x e x Example 52 The special case p = and ψ = γ produces 337: 52 + e x + e x q dx = ψq + γ 53 Example 53 The evaluation of 336: comes directly from 5 + e x p + e x q dx = ψq ψq p

11 DIGAMMA FUNCTION 9 54 Proposition 54 Let p, q R Then This appears as 337 in [2] e pt e qt e t dt = ψq ψp, Proof Mae the change of variables x = e t in 3 Example 55 The evaluation 54 can also be written as e t p e t q e t dt = ψq ψp, Example 56 The special case p = and q = ν is e νt e t dt = ψ ν ψ, and using ψ = γ and ψ ν = ψν + π cot πν, yields the form 57 as it appears in e νt e t dt = ψν + γ + π cot πν, Example 57 Another special case of 54 is 336, that corresponds to p = : e px e qx e rx dx = esx r s e t e qt e t dt = ψq + γ Example 58 The evaluation 33: ψ r q r s ψ r p, r s follows directly from 54 by the change of variables t = r sx 5 Example 59 The evaluation of 332: ψ a x b x c x d x dx = lnc lnd lnc lnb lnc lnd ψ is proved by simply writing the exponentials in natural base lnc lna, ln c lnd Example 5 The formula 33 had a sign error in the sixth edition of [2] It appears as e px e qx π pπ 5 dx = + e p+qx p + q cot p + q It should be 52 The value 59 yields 53 e px e qx π pπ dx = e p+qx p + q cot p + q e px e qx dx = e p+qx p + q q p ψ ψ, p + q p + q

12 2 L MEDINA AND V MOLL and the trigonometric answer follows from 6 This has been corrected in the current edition of [2] 54 Example 5 The evaluation of 3322: e ax e bx e px e x dx = ψp + a + ψp + b ψp + a + b ψp follows directly from 3 Indeed, the change of variables t = e x gives 55 I = and now split them as 56 I = and use 3 to conclude 6 t p t p+a t t p t a t b + t a+b t dt t p+b t p+a+b dt dt t 6 A singular example The example discussed in this section is e µx dx b e x = bµ π cotπµ, that appears as 338 in [2] In the case b > this has to be modified in its presentation to avoid the singularity x = lnb The case b < was discussed in [4] In order to reduce the integral to a previous example, we let t = e x to obtain 62 The change of variables t = by yields 63 e µx dx b e x = e µx dx = bµ b e x t µ dt b t y µ dy y Now separate the range of integration into [, ] and [, Then mae the change of variables y = /z in the second part This produces 64 e µx dx = bµ b e x z µ z µ dz z This last integral has been evaluated as cotπµ in 39 7 An integral with a fae parameter The example considered in this section is 3234: x q 7 ax x q dx = π a x a q cotπq

13 DIGAMMA FUNCTION 2 We show that the parameter a is fae, in the sense that it can be easily scaled out of the formula The integral is written as lim Iǫ where ǫ Iǫ = = x q ax ǫ x q a x ǫ dx x q dx ax ǫ x q dx a x ǫ Mae the change of variables t = ax in the first integral and x = at in the second one to produce a Iǫ = a q t q dt /a t q dt a q+ǫ t ǫ t ǫ, and then let ǫ to produce x q a ax x q dx = a q a x t q dt t /a t q dt t Differentiation with respect to the parameter a, shows that the expression in parenthesis is independent of a It is now evaluated by using a = to obtain x q ax x q dx = a q t q t q dt a x t The evaluation 3 now yields x q ax x q dx a x Formula 7 has been established = a q ψ q ψq = a q π cotπq 8 The derivative of ψ In a future publication we will discuss the evaluation of definite integrals in terms of the polygamma function n d 8 PolyGamma[n, x] := ψx dx In this section, we simply describe some integrals in [2] that comes from direct differentiation of the examples described above Example 8 Differentiating 3 with respect to the parameter p produces 4254: 82 x p lnx x dx = ψ p

14 22 L MEDINA AND V MOLL Example 82 The change of variables x = t q in 82, followed by the change of parameter p p q yields 4254: t p lnt 83 t q dx = p q 2 ψ q 84 Example 83 Replace q by 2q and p by q in 83 to produce t q lnt dt = t2q 4q 2 2ψ To evaluate this last term, differentiate the logarithm of the identity 85 Γ2x = 22x π ΓxΓx + 2, to obtain 86 2ψ2x = 2 ln2 + ψx + ψx + 2 One more differentitation produces 87 4ψ 2x = ψ x + ψ x + 2 The value x = 2 gives 88 ψ 2 = 3ψ = π2 2 Therefore we obtain 42546: 89 x q lnx x 2q dx = π2 8q 2 Example 84 Differentiating 32 n-times with respect to the parameter p produces 4275: 8 ln n x xp dx x q = p ψn qn+ q 9 A family of logarithmic integrals Several of the integrals appearing in [2] are particular examples of the family evaluated in the next proposition 9 92 Proposition 9 Let a, b R + Then x a x b = ΓaΓb Γa + b Proof Differentiate the identity x a x b dx = ΓaΓb Γa + b ψa ψa + b with respect to the parameter a and recall that Γ x = ψxγx

15 DIGAMMA FUNCTION 23 The next corollary appears as 4253 in [2] 93 Corollary 92 Let a, b, c R + Then x a x c b = Γa/cΓb c 2 Γa/c + b Proof Let t = x c in the integral 9 a a ψ ψ c c + b Example 93 The formula in the previous corollary also appears as 4256 in the form x µ dx 94 ln x n xn = µ n m n 2 B n, m [ µ + m µ ] ψ ψ n n n 95 Example 94 The integral x 2n lnx dx = x 2n x 2 /2 x 2 that appears as 424 in [2], corresponds to a = 2n +, b = 2 and c = 2 in 93 Therefore x 2n lnx 96 dx = Γn + 2 Γ 2 [ ψn + x 2 4Γn + 2 ψn + ] Using 7, and 4 yields x 2n 2n lnx 97 dx = n π x 2 This is n+ 2n = Example 95 The integral in 4242 states that x 2n+ lnx dx = ln 2 + x 2 Writing the integral as 99 I = 2n!! 2n +!! ln 2 2n+ = x 2n+ x 2 /2 we see that is corresponds to the case a = 2n + 2, b = 2, c = 2 in 93 Therefore 9 I = Γn + Γ 2 4Γn [ ψn + ψn ] Using 7, and 4 yields 9 x 2n+ ln x x 2 dx = This is equivalent to n n + 2n+ n ln 2 + 2n+ =

16 24 L MEDINA AND V MOLL 92 Example 96 The integral 4243 in [2] states that x 2n x 2 = To evaluate the integral, we write it as 93 I = 2n!! 2n + 2!! π 2 2n = x 2n x 2 /2 2n + 2 ln 2 and we see that is corresponds to the case a = 2n+, b = 3 2, c = 2 in 93 Therefore 94 I = Γn + 2 Γ3 2 4Γn + 2 [ ψn + 2 ψn + 2] Using 7, and 4 yields 95 x 2n 2n x 2 n π = 2 2n+2 ln 2 + 2n n + 2n This is equivalent to Example 97 The integral 4244 in [2] states that x 2n+ x 2 = 2n!! ln 2 + 2n + 3!! To evaluate the integral, we write it as 97 I = 2n+ = x 2n+ x 2 /2 = 2n + 3 and we see that is corresponds to the case a = 2n+2, b = 3 2, c = 2 in 93 Therefore 98 I = Γn + Γ3 2 4Γn [ ψn + ψn ] Using 7, and 4 yields 99 x 2n+ x 2 = This is equivalent to n+ n + n + 2 2n+3 n+ Example 98 The integral 4245 in [2] states that ln 2 2n+ 2n = lnx x 2 2n 2n!! π dx = [ψn + + γ + ln 4] 4 2n!! To evaluate the integral, we write it as 92 I = x 2 n 2

17 DIGAMMA FUNCTION 25 and we see that is corresponds to the case a =, b = n + 2, c = 2 in 93 Therefore 922 I = Γn + 2 Γ 2 4Γn + [ ψ 2 ψn + 2 ] Using 7, and 4 yields 2n 923 x 2 n 2 n π = 2 ln n+2 n = This is equivalent to 92 This integral also appears as Example 99 The case n = in 97 yields This appears as 4247 in [2] = π ln 2 x 2 2 Example 9 Formula 4248 states that 925 x = ln 2 2 x 2 To evaluate this, let t = /x to obtain 926 I = t t 2 /2 lntdt This corresponds to the case a = 2, b = 2, c = 2 in 93 Therefore 927 I = ΓΓ 2 4 Γ 3 2 [ ψ ψ 3 2 ], and the value ln 2 comes from and 4 Example 9 The case n = in 95 produces 928 x2 = π 2 ln This appears as 4249 in [2] 929 Example 92 The case n = in 99 produces This appears as 424 in [2] 93 x x 2 = 3 ln2 4 9 Example 93 Entry 424 states that To evaluate the integral, write it as 93 I = 2π x x2 = 8 Γ2 4 x /2 x 2 /2

18 26 L MEDINA AND V MOLL and this corresponds to the case a = 2, b = 2, c = 2 in 93 Therefore 932 I = Γ 4 Γ 2 [ ψ 4Γ 4 ψ 3 ] The stated form comes from using 9 and Example 94 The identity x lnx x 4 dx = π 8 ln 2 appears as 4243 in [2] To evaluate it, we write it as 934 I = x x 4 /2 that corresponds to a = 2, b = 2, c = 4 in 93 Therefore, 935 I = 6 Γ2 2 [ ψ 2 ψ] The values ψ = γ and ψ 2 = γ 2 ln2 gives the result 936 Example 95 The verification of 4244: x x2 = 2 8 Γ3 is achieved by using 93 with a = 2 3, b = 3 and c = 2 to obtain 937 I = Γ2 [ 3 ψ 4Γ 3 ψ 2 ] Using 9 and 6 produces the stated result Example 96 The usual application of 93 shows that is 3 = 2π x 3 9 ψ 3 + γ, 3 where we have used Γ 3 Γ2 3 = 2π/ 3 It remains to evaluate ψ 3 The identity 6 gives 939 ψ 3 ψ 2 3 = π 3 To obtain a second relation among these quantities, we start from the identity 94 Γ3x = 33x /2 ΓxΓx + 2π 3 Γx that follows directly from 6, and differentiate logarithmically to obtain 94 ψ3x = ln 3 + ψx + ψx ψx The special case x = 3 yields 942 ψ 3 + ψ 2 3 = 2γ 3 ln3 3

19 DIGAMMA FUNCTION 27 We conclude that 943 ψ 3 = γ 3 2 ln 3 π 2 3 and 944 ψ 2 3 = γ 3 2 ln 3 + π 2 3 This gives 945 as stated in = π x 3 3 ln 3 + π 3 3, 3 Example 97 The evaluation of 42443: 3 x 3 = π 3 3 ln 3 π 3 3 proceeds as in the previous example The integral is identified as 947 I = 9 Γ2 3 Γ 3 [ ψ γ ] The value 944 gives the rest 948 Example 98 The change of variables t = x 4 yields x p x 4 = 6, t p 3/4 t /2 lntdt The last integral is evaluated using 93 with a = p+ 4, b = 2 and c = to obtain x p π 949 = Γ p+ 4 [ ] p + p + 3 x 4 6 Γ p+3 ψ ψ 4 4 The special case p = 4n + yields 95 x 4n+ x 4 = The special case p = 4n + yields x 4n+ x 4 = 4 π 6n! Γn + 2 [ ψn + 2 ψn + ] π Γn + 2 [ ψn + 6n! 2 ψn + ] Using 7, and 4 yields 4245 in the form x 4n+ 95 = π 2n 2n n ln 2 + x 4 The special case p = 4n + 3 yields x 4n+3 x 4 = 2 2n+3 = π n! 6Γn [ ψn + ψn ]

20 28 L MEDINA AND V MOLL Using 7, and 4 yields in the form x 4n+3 2 2n = ln 2 + x n + 2n n 2n+ = Example 99 The change of variables t = x 2n produces n x 2n = 4n 2 Then 93 with a = 2n, b = n 954 n = Γ 2n Γ n x 2n t 2n t n lntdt and c = give the value [ ψ 4n 2 Γ 2n 2n Using 7 and 4 to obtain 955 n = π B 2n, 2n x 2n 8 n 2 sin π 2n This is 4247 in [2] Example 92 The change of variables t = x 2 gives Using 93 we obtain 956 n x n x 2 = 4 x n x 2 = Γ 2n Γ n ψ 2n] t 2n t n lntdt 2n [ ψ Γ 2n 2n Proceeding as in the previous example, we obtain 957 n x n x 2 = π B 2n, 2n 8 sin π 2n This is in [2] ψ 2n] Some integrals in [2] have the form of the Corollary 92 after an elementary change of variables 958 Example 92 Formula in [2] states that x a ln xdx = ψa + + γ a This follows directly from 93 by the change of variables x x The same is true for 42933: 959 x a x b ln xdx = Ba, b[ψb ψa + b]

21 DIGAMMA FUNCTION Example 922 The change of variables t = e x gives xe x e 2x n 2 dx = t 2 n 2 lntdt This latter integral is evaluated using 93 as πγn I = ψ 4n! 2 ψn + Using 7 and we obtain 2n 962 xe x e 2x n 2 n π n dx = 2 2n+2 2 ln2 + This appears as 3457 in [2] An announcement There are many integrals in [2] that contain the term + x in the denominator, instead of the term x seen, for instance, in Section 3 The evaluation of these integrals can be obtained using the incomplete beta function, defined by βa := x a dx + x as it appears in 8372 This function is related to the digamma function by the identity 2 βa = [ a + a ] ψ ψ These evaluations will be reported in [5] One more family We conclude this collection with a two-parameter family of integrals Proposition Let a, b R + Then e xa dx + x b x = γ a, independently of b Proof Write the integral as 2 e xa e xb dx x + = e xb dx + x b x The first integral is a bγ/ab according to 2 The change of variables t = x b converts the second one into 3 e t dt b + t t = γ b, according to 26 The formula has been established

22 3 L MEDINA AND V MOLL Example 2 The case a = 2 n and b = 2 n+ gives 3475: dx 4 exp x 2n + x 2n+ x = γ 2 n Example 3 The case a = 2 n and b = 2 gives 34752: 5 exp x 2n dx + x 2 x = γ 2 n Example 4 The case a = 2 and b = 2 gives 3467: 6 e x2 dx + x 2 x = γ 2 7 Example 5 Finally, the change of variables t = ax yields e px dx + a 2 x 2 x = e pt/a e t dt+ t e t + t 2 dt t The first integral is ln a p according to 25, the second one is γ This gives the evaluation of 34423: 8 e px dx + a 2 x 2 x = γ + ln a p References [] G Boros and V Moll Irresistible Integrals Cambridge University Press, New Yor, st edition, 24 [2] I S Gradshteyn and I M Ryzhi Table of Integrals, Series, and Products Edited by A Jeffrey and D Zwillinger Academic Press, New Yor, 7th edition, 27 [3] V Moll The integrals in Gradshteyn and Ryzhi Part 4: The gamma function Scientia, 5:37 46, 27 [4] V Moll The integrals in Gradshteyn and Ryzhi Part 6: The beta function Scientia, 6:9 24, 28 [5] V Moll The integrals in Gradshteyn and Ryzhi Part : The incomplete beta function Scientia, to appear Department of Mathematics, Tulane University, New Orleans, LA 78 address: lmedina@mathtulaneedu Department of Mathematics, Tulane University, New Orleans, LA 78 address: vhm@mathtulaneedu Received 29 27, revised 8 28