A Closed Form for ∫₀^∞ (x coth x − 1) Li₂(e^(−x)) / x dx, Derived Two Ways

The integral below was posted on Math Stack Exchange together with a claimed closed form:

∫₀^∞ (x coth x − 1) Li₂(e^(−x)) / x dx = (π²/6) ln(A¹² / (4π)) − (3/8) ζ(3)

Here A is the Glaisher–Kinkelin constant, ζ is the Riemann zeta function and Li₂ is the dilogarithm. The value is approximately 0.296077216103125, a figure given in the accepted answer. The asker had a working derivation, but it used a Laplace transform, Binet's second formula for ln Γ, and two identities from C.I. Vălean's More (Almost) Impossible Integrals, Sums, and Series (2023). The question was whether a more direct route exists.

One does. Below are both derivations, the parametric generalisation behind the result, and the points where the formulas need care.

Route one: a Laplace transform of x coth x − 1

The original derivation starts with the standard transform ∫₀^∞ sin(at) e^(−st) / t dt = arctan(a/s). Dividing by e^(2πa) − 1 and integrating over a from 0 to ∞ produces the integral term in Binet's second formula for ln Γ(z). The result is a Laplace transform of the kernel:

∫₀^∞ (x coth x − 1) e^(−sx) / x² dx = 2 ln Γ(s/2) + s − (s − 1) ln(s/2) − ln(2π),   Re(s) > 0

Differentiating with respect to s removes one power of x and brings in the digamma function:

∫₀^∞ (x coth x − 1) e^(−sx) / x dx = ln(s/2) − 1/s − ψ(s/2),   Re(s) > 0

The series Li₂(e^(−x)) = Σ e^(−sx)/s² explains the next step. Divide by s² and sum over s = 1, 2, 3, … The left side becomes the target integral. The right side splits into −ζ′(2) from the logarithm, −(π²/6) ln 2, −ζ(3) from the 1/s term, and one series that still needs work: −Σ ψ(s/2)/s².

Harmonic numbers and the digamma series

The remaining sum is where harmonic numbers come in. For a positive integer n, Vălean's book gives ψ(n/2) in terms of the harmonic number Hₙ = Σ 1/k and the skew-harmonic number H̄ₙ = Σ (−1)^(k−1)/k:

ψ(n/2) = −γ − ln 2 − 2/n − (−1)^(n−1) ln 2 + Hₙ + (−1)^(n−1) H̄ₙ

The alternating piece comes from a second identity, ∫₀¹ x^(n−1) ln(1 + x) dx. Writing that integral in terms of H̄ₙ, dividing by n and summing gives Σ (−1)^(n−1) H̄ₙ / n² = (π²/4) ln 2 − (5/8) ζ(3). Combined with the digamma identity, this gives:

Σₙ ψ(n/2) / n² = −(π²/6) γ − (5/8) ζ(3)

Substituting back gives (π²/6)(γ − ln 2) − ζ′(2) − (3/8) ζ(3). The last step uses ζ′(2) = (π²/6)(γ + ln(2π) − 12 ln A). The γ terms cancel, and the logarithms combine into ln(A¹²/(4π)).

Route two: integrate by parts first

The accepted answer avoids the digamma series entirely. The key observation is that (x coth x − 1)/x = coth x − 1/x, which is the derivative of ln(sinh x / x). Also, d/dx Li₂(e^(−x)) = ln(1 − e^(−x)). Integrating by parts therefore gives:

I = −∫₀^∞ ln(sinh x / x) ln(1 − e^(−x)) dx

Next, rewrite −ln(sinh x / x) as ln(2x) − x − ln(1 − e^(−2x)) and handle the three pieces separately:

• The ln(2x) piece is a Mellin-type integral. It equals −d/ds [2^(s−1) Γ(s) ζ(s+1)] at s = 1, which is where ζ′(2), γ and ln 2 appear.
• The −x piece expands term by term to Σ 1/k³ = ζ(3).
• The last piece, after substituting t = e^(−x), becomes −K, where K = ∫₀¹ ln(1 − t) ln(1 − t²) / t dt.

K follows from two classical integrals: ∫₀¹ ln²(1 − t)/t dt = 2ζ(3) and ∫₀¹ ln²(1 + t)/t dt = ζ(3)/4. Expanding the product of logarithms through ln(1 − t²) = ln(1 − t) + ln(1 + t) gives K = (11/8) ζ(3). The boxed result matches route one exactly: (π²/6)(γ − ln 2) − ζ′(2) − (3/8) ζ(3).

The parametric family behind the result

The asker also stated a generalisation with Li₂(z e^(−x)) in the integrand, valid for real z in (−1, 1) excluding 0. Its closed form is long. It contains ln³ 2, γ Li₂(z), several products of ln(1 − z), the trilogarithms Li₃(1 − z), Li₃(z²), Li₃((1 − z)/2), Li₃(z/(z − 1)) and Li₃(2z/(z − 1)), and a final term involving the derivative of Li_s(z) with respect to the order s at s = 2. That last term is the parametric counterpart of ζ′(2) in the special case.

Setting z to −1 gives a companion result:

∫₀^∞ (x coth x − 1) Li₂(−e^(−x)) / x dx = (π²/12) ln(8π / A¹²) − (3/8) ζ(3)

Combining this with the original integral gives two more closed forms. The version with Li₂(e^(−2x)) evaluates to (π²/6) ln(A¹²/(2π)) − (3/2) ζ(3). The version with the Legendre chi function χ₂(e^(−x)) evaluates to (π²/8) ln(A¹² / (4π · 2^(1/3))), and its ζ(3) terms cancel completely.

What to watch for before reusing these formulas

The general formula is stated on the open interval (−1, 1), but both z = 1 and z = −1 lie on its boundary. At z = 1, terms such as ln(1 − z) diverge on their own. The original integral is therefore a limit in which those divergences cancel, not a direct substitution. If you evaluate the general formula numerically near the endpoints, expect cancellation error.

For negative z, ln(2z) is not real. Any software implementation needs consistent branch choices for the logarithm and the trilogarithms. Check this against a direct quadrature at a few interior points before relying on it.

Route one also swaps an infinite sum with an integral and depends on two identities from a single book. It is correct, but it is harder to audit than route two. Route two needs only integration by parts, a Mellin transform and two standard log-squared integrals. That makes it the better template for related integrals with a different polylogarithm against the same kernel. In either case, check the final value against 0.296077216103125 to catch sign errors in ζ′(2), which are easy to make.

This article draws on Closed form of $\int_0^\infty\frac{(x\coth(x)-1)\operatorname{Li}_2(e^{-x})}{x}\,\mathrm dx$ and an answer to it, on the Stack Exchange network, licensed under CC BY-SA 4.0.