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Gauss–Jacobi regularization for x to the minus one half times log x

Statement

The improper integral is ∫01x−1/2log⁡x dx=−4. For example, Finite endpoint power substitution followed by Gauss–Jacobi quadrature applies with n=1, α=−1/2, β=0, ma=1, mb=0, h(x)=1, qa=7, and qb=4. The transformed functions are C2 and Error bound after endpoint power regularization gives an explicit error bound C1(∥FL′′∥∞+∥FR′′∥∞) for the returned two-node sum.

Facts & Assumptions

Given: The integrand f(x)=x−1/2log⁡x on (0,1] and the specified endpoint-regularization inputs.

[F1]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: log⁡ is differentiable on (0,∞) with derivative 1/x.

[F2]

A sufficiently high power regularizes finite logarithmic endpoint factors: tδP(log⁡t) extends to Cr([0,1]) when δ>r and P is a polynomial.

[F3]

Finite endpoint power substitution followed by Gauss–Jacobi quadrature: the left and right substitution orders qa,qb and the shifted Gauss–Jacobi sums define the stated finite output.

[F4]

Error bound after endpoint power regularization: when the substituted functions are C2n, the improper integral exists and the error is bounded by the displayed derivative constant.

Proof

1.1F1F2F5F6algebra

A primitive on (0,1] is P(x)=2xlog⁡x−4x by [F1] and [F6]. The endpoint term xlog⁡x tends to zero by the power-regularity lemma [F2] with r=0, δ=1/2, and polynomial z. Hence P(x)→0 as x↓0. Applying [F5] on [ε,1] and taking ε↓0 gives the improper integral P(1)−0=−4.

1.2F2F3givenalgebra

For the left substitution, δL=qa(α+1)−1=7/2−1=5/2>2; its transformed function is a constant times t5/2 times a polynomial of degree one in log⁡t and a smooth factor bounded with two continuous derivatives. For the right substitution, δR=qb(β+1)−1=3>2 and the remaining factors are smooth on [0,1]. Therefore both transformed functions are C2 by [F2], and [F3] applies.

2.1F3F4step 1.1step 1.2∎

Apply [F4] with n=1 to the functions in step 1.2. Its error constant is C1 and the output is the stated sum, so the claimed error bound follows. The value −4 from step 1.1 is the integral being approximated.

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