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By Cantwait4colleg (Cantwait4colleg) on Monday, May 10, 2004 - 10:40 pm: Edit |

Stupid question...

Integral of (1/x)dx from 0 to 1...

By Miseryxsignals (Miseryxsignals) on Monday, May 10, 2004 - 10:52 pm: Edit |

integral of 1/x = lnx.

soo... ln(1) - ln(0). And you cant do the ln(0), Sorry.

By Athlonmj (Athlonmj) on Monday, May 10, 2004 - 10:54 pm: Edit |

Yes you can, it's an improper integral so substitute a for 0 and take the limit as it approaches 0.

By Efilsiertaeht (Efilsiertaeht) on Monday, May 10, 2004 - 11:05 pm: Edit |

approaches 0, not infinity.

By Sandwraith (Sandwraith) on Monday, May 10, 2004 - 11:05 pm: Edit |

either way the point is that the value of the integral is not a finite answer

By Mike105 (Mike105) on Monday, May 10, 2004 - 11:07 pm: Edit |

ò_{0 }^{1}(1/x)dx = ¥

By Miseryxsignals (Miseryxsignals) on Monday, May 10, 2004 - 11:13 pm: Edit |

haha oops. Shows what a year w/o math can do to you. SOrry about that, thanks for catching my mistake!

By Ubercollegeman (Ubercollegeman) on Monday, May 10, 2004 - 11:40 pm: Edit |

That is an improper integral. If you want the full way to do it..

lim a->0+ ln(1) - ln(a) =

lim a->0+ 0 - lna = -(-infty) = infty.

^not good mathematical notation, but whatever!

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