Apr 29th 2011, 02:42 PM #11 Member   Join Date: Nov 2010 Posts: 40 Let me work out this and see whether I am correct: $\int_{-b}^{-a}\frac {dx}{|x|} = sgn(-a)ln|-a|-sgn(-b)ln|-b| = -ln|-a|-(-ln|-b|) = ln\left | \frac {-b}{-a}\right | = ln\left | \frac {b}{a}\right |$ And $\int_{-b}^{a}\frac {dx}{|x|} = sgn(a)ln|a|-sgn(-b)ln|-b| = ln|a|-(-ln|-b|) = ln|a(-b)|= ln(ab)$ Am I correct? Thanks Alan Last edited by yungman; Apr 29th 2011 at 02:55 PM.
Apr 29th 2011, 04:40 PM   #12

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 Originally Posted by yungman Let me work out this and see whether I am correct: $\int_{-b}^{-a}\frac {dx}{|x|} = sgn(-a)ln|-a|-sgn(-b)ln|-b| = -ln|-a|-(-ln|-b|) = ln\left | \frac {-b}{-a}\right | = ln\left | \frac {b}{a}\right |$ And $\int_{-b}^{a}\frac {dx}{|x|} = sgn(a)ln|a|-sgn(-b)ln|-b| = ln|a|-(-ln|-b|) = ln|a(-b)|= ln(ab)$ Am I correct? Thanks Alan
Looks good to me.

-Dan
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 Apr 29th 2011, 05:05 PM #13 Member   Join Date: Nov 2010 Posts: 40 Thanks so much. You are of big help Dan. Alan

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