设函数

阿甘

npub13zyg3zysfylqc6nwfgj2uvce5rtlck2u50vwtjhpn92wzyusprfsdl2rce

hex

dc8758abed33049f011ec6b216383db837e381946d75b4b83894d04894b33e4f

nevent

nevent1qqsdep6c40knxpylqy0vdvsk8q7msdlrsx2x6ad5hquff5zgjjenuncprpmhxue69uhhyetvv9ujuem4d36kwatvw5hx6mm9qgsg3zyg3zgyj0svdfhy5f9wxvv6p4lut9w28k89etsej48pzwgq35cryy8nx

Kind-1 (TextNote)

2026-03-02T12:20:45Z

↳ Reply to 阿甘 (npub13zyg3zysfylqc6nwfgj2uvce5rtlck2u50vwtjhpn92wzyusprfsdl2rce)

nostr:npub1gcd3atn3dv9l9hjrql2jcvecczgppemsg4n2lw3f7vad2s5f5ydqu7mht7 瑟瑟发抖

设函数

f(x) = \frac{\sin(\pi x)}{x^2 - 2x + 2} + e^{-x^2}\ln\left(1 + \frac{1}{1+x^2}\right)

定义

I = \lim_{n\to\infty} \int_{-n}^{n} \left( f(x) - \frac12 f''(x) \right) \arctan\left(e^x\right) dx

再令

J = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \int_{0}^{1} \frac{x^{k-1}}{\sqrt{1 - x^2}} dx

求

K = I + J

并给出 K的精确值。

Raw JSON

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  "content": "设函数\n \nf(x) = \\frac{\\sin(\\pi x)}{x^2 - 2x + 2} + e^{-x^2}\\ln\\left(1 + \\frac{1}{1+x^2}\\right)\n \n定义\n \nI = \\lim_{n\\to\\infty} \\int_{-n}^{n} \\left( f(x) - \\frac12 f''(x) \\right) \\arctan\\left(e^x\\right) dx\n\n再令\n \nJ = \\sum_{k=1}^{\\infty} \\frac{(-1)^{k+1}}{k} \\int_{0}^{1} \\frac{x^{k-1}}{\\sqrt{1 - x^2}} dx\n \n求\n\nK = I + J\n\n并给出 K的精确值。",
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