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DefiniteIntegral is a first-order rule, and is NaN where an endpoint is singular

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难度
4/5
预计耗时
3-5 天
新手友好度
48/100
Issue 类型
缺陷
描述清晰度
基本清楚
活跃度
活跃
技术栈
csharp
领域
backend

调研方向

Start at Entity.DefiniteIntegral(x, a, b, steps) and trace its implementation and the MCP server (amcli mcp). Check x^2 on [0, 1] and the singular-endpoint cases sin(x)/x and ln(x) at the stated intervals and step counts. Done means endpoint singularities no longer produce NaN, integration accuracy improves, and MCP output reports only digits agreed between runs.

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描述

Entity.DefiniteIntegral(x, a, b, steps) is a rectangle rule. On x^2 over [0, 1] its error is 1.67e-4 at 1000 steps, 8.3e-5 at 2000 and 4.2e-5 at 4000: it halves with each doubling. The value is still printed to a hundred digits, nearly all of them wrong. Simpson's rule costs the same evaluations, is exact on this integrand, and its error on a smooth one falls sixteenfold per doubling.

It also evaluates both endpoints, so a convergent integral with a singular endpoint has no value. sin(x)/x over [0, 1], which is Si(1) = 0.946083..., and ln(x) over [0, 1], which is -1, are both NaN. A midpoint or Gauss–Legendre rule never evaluates an endpoint.

The MCP server (amcli mcp) runs it at two step counts and reports only the digits the two agree on.

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