Triple integral of log z

I = ∫₁ᵉ ∫₁^{log y} ∫₁^{eˣ} log z dz dx dy

The integration order is z (innermost), then x, then y.

z-integration

∫ log z dz = z log z − z, so

∫₁^{eˣ} log z dz = [z log z − z]₁^{eˣ} = (x eˣ − eˣ) − (0 − 1) = (x − 1)eˣ + 1

x-integration (1 to log y)

∫ (x − 1)eˣ dx = (x − 2)eˣ, so

∫₁^{log y} [(x−1)eˣ + 1] dx = [(x − 2)eˣ + x]₁^{log y}
  = (log y − 2) y + log y − (−e + 1)
  = y log y − 2y + log y + e − 1

y-integration (1 to e)

∫₁ᵉ y log y dy = [ (y²/2) log y − y²/4 ]₁ᵉ = e²/2 − e²/4 + 1/4 = e²/4 + 1/4
∫₁ᵉ 2y dy      = e² − 1
∫₁ᵉ log y dy   = [y log y − y]₁ᵉ = 0 + 1 = 1
∫₁ᵉ (e − 1) dy = (e − 1)²

I = (e²/4 + 1/4) − (e² − 1) + 1 + (e − 1)²
  = e²/4 + 13/4 − 2e
  = (e² − 8e + 13)/4

Answer: I = (e² − 8e + 13)/4 ≈ −0.34.

The value is negative because for 1 < y < e the upper limit log y of x is below the lower limit 1, so that part is integrated backwards. The integral is evaluated exactly as printed.