Binet's second formula, Hermite's generalization, and two related identities

1.1 Summary of results

The following two results are due to Legendre [1]. Let a ∈ R ⧹ { 0 } , then

The left-hand sides of these two equations vanish when a = 0 . In the limit that a → 0 , the right-hand sides also vanish (as can be shown by taking their Taylor expansions), and so these formulas are in fact valid for all a ∈ R if the right-hand sides are extended by continuity. Equations (1) and (2) appear as Exercises 41 and 42 on page 71 of Watson’s contour-integration book [2]. Watson attributes these results to Legendre, and moreover, he states that they hold for all real a . An alternative method to derive (1) and (2), based on infinite series and term-by-term integration, is illustrated in Examples 1 and 2 of Section 176 of Bromwich’s infinite-series book [3].

The analogous integrals with cosines appearing in the integrand instead of sines can be expressed (cf. Section 2.5.34 of [4]) in terms of the “digamma” function ψ (for which, see, for instance, Section 12.3 of [5]). Here, however, we shall express them in the following forms:

These formulas are valid for all a ∈ R ⧹ { 0 } , and in the limit a → 0 , they also agree. The integrands on the left-hand sides of equations (1) and (3) extend continuously to x = 0 , and the integrand on the right-hand side of equation (3) extends continuously to the whole interval [ 0 , 1 ] ; this is, indeed, why it is presented in the form shown above. See the penultimate equality of Section 2.1 for the “improper” Riemann integral version of the result. Similarly, the integrand on the right-hand side of equation (4) extends continuously to the whole interval [ 0 , 1 ] , with the integrand having a finite limit as x → 1 2 . The Cauchy principal value form of this result can be deduced from the last equation of Section 2.2. A brief derivation of equations (1)–(4) is presented in Section 2.

Binet’s second formula [6] for log Γ ( z ) (see Sections 12.31 and 12.32 of [5] for Binet’s first and second formulas) is stated as follows: let Re ( z ) > 0 , then

In Sections 177 and 180 of [3], equation (1) is elegantly used to derive Binet’s second formula. Here, we shall show that, when the same technique is applied to equation (2), the result is Hermite’s formula [7] for log Γ z + 1 2 :

By also applying the same analysis to equations (3) and (4), we shall establish the following theorem:

The square root is defined to be its “principal value.” Let z be in the open first/fourth quadrant; so are z + 1 and z / ( z + 1 ) . Then, z 2 and ( z + 1 ) 2 are both in the open upper/lower half-plane, as are x 2 + ( z + 1 ) 2 and x 2 + z 2 (for real x ). Thus, h ( x , z ) ≔ [ x 2 + ( z + 1 ) 2 ] / ( x 2 + z 2 ) has argument in ( − π , π ) , and this is also true if z is real and positive. Thus, the principal value of h ( x , z ) is holomorphic in z (for real nonnegative x ) and belongs to the same right-hand quadrant as z . Hence, [ z / ( z + 1 ) ] h ( x , z ) does not take values on the negative real axis, and so the principal value of the logarithm is also holomorphic on the right-hand half-plane.

We shall also show that equations (7) and (8) can be used to deduce the following results:

Binet’s second formula (5) involves an integral of an arctangent (in general with a complex argument). In the restricted case of a real argument, the arctangent function can be expressed as the imaginary part of a complex logarithm. Thus, if we replace the arctangent function in Binet’s second formula by a complex logarithm, then a natural problem is to evaluate the real part of this integral. We shall address this problem by using the two identities (9) and (10). Indeed, we can take z to be a positive real number, and by using equations (9) and (10) along with Binet’s second formula (5) and Hermite’s generalization (6), we have the general result as follows:

1.2 Review of Legendre’s work

For historical interest, let us briefly outline the method that Legendre used to derive (1) and (2). In Section 44, page 186 of Volume 2 of [1], Legendre provides several integration identities involving sines and cosines. The two integrals involving sines are as follows (we slightly adapt the notation from Legendre’s form): let ∣ b ∣ < π , then

The integrand in equation (13) extends continuously to x = 0 . In both (13) and (14), ∣ b ∣ < π ensures that the right-hand side has a positive denominator and the integral on the left-hand side converges. Let b = π − ε , where 0 < ε < 1 2 π . The left-hand side of equation (13) becomes

For the integral on the right-hand side of the above expression, the integrand is bounded by an integrable function, since, for 0 < ε < 1 2 π ,

The dominated convergence theorem [8] may thus be applied, and so the integral and the limit ε → 0 + may be interchanged. After taking the limit ε → 0 + on the right-hand side of equation (13) as well, we obtain

The result above appears on page 189 of [1]. Rearranging this equation and simplifying it leads to equation (1). Applying the same procedure to equation (14) leads to the following result:

Taking the limit ε → 0 + on the right-hand side of equation (14) and the equation above, we obtain

Rearranging this equation and simplifying it leads to equation (2). In Section 44, page 186 of [1], Legendre also has two integral identities involving cosines. However, for each identity, the relative signs in the integrand are the same for the numerator and denominator. Thus, the identities obtained by Legendre involving cosine do not lead to the integrals expressed in (3) and (4). In Section 2, we present a contour-integral-based derivation of equations (1)–(4).

2.1 First set of integrals

To prove equations (1) and (3), let a ∈ R ⧹ { 0 } and consider the contour integral

The closed contour C is traversed in an anticlockwise direction and has vertices in the complex plane at ( ε , 0 ) , ( R , 0 ) , ( R , 1 ) , ( ε , 1 ) , ( 0 , 1 − ε ) , and ( 0 , ε ) , where the contour is indented, for instance, by clockwise circular quadrants of radius ε at ( 0 , 0 ) and ( 0 , 1 ) . By Cauchy’s theorem, J = 0 . There are six contributions to the integral, and they are given as follows:

The indentations about ( 0 , 1 ) and ( 0 , 0 ) have contributions, respectively, given by

Finally, the remaining contribution to the contour integral J is

Now take the limits R → ∞ and ε → 0 + . As R → ∞ , I 2 ( R ) → 0 . Combine all these results, set the real and imaginary parts of J equal to zero and then simplify to obtain

The second line is the result in equation (1). To write the result in the first line in the more convenient form (3), we use the fact that lim ε → 0 + ∫ ε 1 − ε cot ( π x ) d x = 0 . Subtracting ( 1 − e − a ) − 1 multiplied by the preceding integral from the first line leads to the identity in equation (3).

2.2 Second set of integrals

To prove equations (2) and (4), consider the contour integral

The closed contour C is traversed in an anticlockwise direction and has vertices at ( 0 , 0 ) , ( R , 0 ) , ( R , 1 ) , ( 0 , 1 ) , 0 , 1 2 + ε , and 0 , 1 2 − ε , where the contour is indented, for instance, by a clockwise semicircle of radius ε about 0 , 1 2 . By Cauchy’s theorem, J = 0 . There are five contributions to the integral, and they are given as follows:

The clockwise indentation about 0 , 1 2 is given by

Finally, the remaining contribution to the contour integral J is the Cauchy principal value integral given as follows:

Now take the limit R → ∞ . As R → ∞ , I 2 ( R ) → 0 . Combining all the above results, we obtain

Next, we set both the real and imaginary parts of J equal to zero and then simplify. For the imaginary part, the result obtained is equation (2). For the real part, however, we note that the integrand in the Cauchy principal value integral is discontinuous at x = 1 2 . As shown in Appendix A, P ∫ 0 1 tan ( π x ) d x = 0 . Thus, by subtracting e − a 2 tan ( π x ) from the integrand of the Cauchy principal value integral above, we ensure that the integrand has a finite limit as x → 1 2 . We can then continuously extend the domain of integration to [0,1] without computing a Cauchy principal value integral. After this modification, the result in equation (4) is obtained.

3.1 Hermite’s generalization of Binet’s second formula for log Γ ( z )

An interesting application of equation (1) is to use it to prove Binet’s second formula [6] for the logarithm of the gamma function. For convenience, we restate this formula here. Let Re ( z ) > 0 , then

The proof is outlined in Sections 177 and 180 of [3]. In this section, we shall apply Bromwich’s method to the other Legendre sine integral (2), and we shall find that the result is Hermite’s formula [7] for log Γ z + 1 2 , which is a special case of his expansion for log Γ ( z + a ) when 0 < a < 1 .

Let Re ( w ) > 0 and multiply both sides of equation (2) by e − a w before integrating over a from zero to infinity as follows:

In this article, all integrals with infinite domains of integration and continuous integrand are absolutely convergent, and thus there is no difficulty in reversing the order of integration in repeated integrals of this kind. This fact was used in obtaining (16). Further discussion can be found in Section 177 of [3]. To evaluate the integral on the left-hand side of the first line in equation (16), we use the integral representations of the logarithm (see Section 6.222, page 116, Example 6 of [5]) and of the digamma function ψ (see Section 12.3, page 247 of [5]). For Re ( w ) > 0 ,

In both cases, the integrand extends continuously to a = 0 . By using equations (17) and (18), the left-hand side of equation (16) can be evaluated. Thus, we find

As a result, we have

Let w = ξ + i η and z = ξ 0 + i η , where ξ 0 > 0 , and integrate both sides of equation (19) with respect to ξ , from ξ = ξ 0 to ξ = ∞ . For the left-hand side, we obtain

Here, we have again reversed the order of integration. The arctangent here is defined on the right-hand half-plane, with value 0 on the real axis (cf. line 11, page 251 of [5]):

where the path of integration is a straight line and u ≠ + i , − i . Thus, the limit of arctan ( x / w ) as ξ → ∞ is 0 ( x being positive real), and its derivative with respect to w is − x / ( x 2 + w 2 ) ; the signs cancel, and thus the ξ integral performed earlier indeed results in arctan ( x / z ) .

For the integral of the right-hand side of (19), we can use the fact that ψ ( w ) = d d w log Γ ( w ) to obtain

Evaluating the anti-derivative at the lower limit of integration is straightforward, and for the upper limit, we rewrite the anti-derivative in a more convenient form, which results in the following expression:

Note that, in the above expression, w = ξ + i η . A corollary of Binet’s first formula (see the last paragraph of page 249 of [5]) is that, if z = x + i y , then, for large x ,

Thus, when x is large, the terms z − 1 2 log z − z + 1 2 log ( 2 π ) furnish an approximate expression for log Γ ( z ) . Hence, using this result (or, equivalently, Stirling’s series for the logarithm of the Gamma function; see page 252 of [5]), as ξ → ∞ , the term on the second line of the right-hand side of (21) tends to zero. The term on the third line of the right-hand side of (21) tends to 1 4 log ( 2 π ) , and hence,

Finally, since (20) is the result for the integral of the left-hand side of (19), and (23) is the result for the integral of the right-hand side of (19), we can equate these two expressions and rearrange to find, for Re ( z ) > 0 ,

The result in (24) corresponds to the specific case a = 1 2 in Hermite’s general formula [7] for log Γ ( z + a ) ; see also the expression below equation (1.10) in [9], where log Γ ( z + a ) , which is written in terms of two integrals, reduces to (24) for the case a = 1 2 .

3.2 Proof of Theorem 1.1, equation (7)

We now apply the procedure from the previous section to the integrals in equation (3). First, we multiply the left-hand side of equation (3) by e − a w ( 1 − e − a ) , where Re ( w ) > 0 , and then integrate over a to obtain

Performing the same procedure on the right-hand side of equation (3) leads to

Note that, as mentioned in Section 1.1, the integrand on the right-hand side of equation (3) extends continuously to the whole interval [ 0 , 1 ] . This is relevant here because it means that we can avoid an improper Riemann integral argument. The integrand in equation (26) also extends continuously to the interval [ 0 , 1 ] .

As in the previous section, we let w = ξ + i η and z = ξ 0 + i η , where ξ 0 > 0 , and integrate over ξ from ξ = ξ 0 to ξ = ∞ . Performing these operations on (25), we find

The above final equality can be obtained by decomposing the integrand into partial fractions and performing the integration over ξ .

Now perform the same procedure on (26) to obtain

The above final equality is again obtained by decomposing the integrand into partial fractions and performing the ξ integration. The integrand on the right-hand side of (28) extends continuously to [ 0 , 1 ] . Interchanging the order of integration in the previous repeated integrals is permissible in all cases since, as in the previous section, the integrals are absolutely convergent. Since the expressions in (25) and (26) are equal, the same is true for the expressions in (27) and (28), and therefore this proves equation (7).

3.3 Proof of Theorem 1.1, equation (8)

We now analyze equation (4). First, we multiply the left-hand side of equation (4) by e − a w ( 1 − e − a ) , where Re ( w ) > 0 , and then integrate over a to obtain

Performing the same procedure on the right-hand side of equation (4) leads to

In (29), let w = ξ + i η and z = ξ 0 + i η , where ξ 0 > 0 , and integrate over ξ from ξ = ξ 0 to ξ = ∞ :