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Convergence of integral

  1. Apr 13, 2012 #1
    I know that for any [itex]a>0[/itex] and [itex]k,t\in\mathbb{R}[/itex], the integral [tex]\int_0^a t^k\; dt[/tex] converges if and only if [itex]k>-1[/itex].

    Is it true that if [itex]k[/itex] is complex then [tex]\displaystyle \int_0^a |t^k| \; dt[/tex] converges if and only if [itex]\text{Re}(k)>-1[/itex] since if [itex]t[/itex] is real, [itex]|t^k|[/itex] does not depend on the imaginary part of [itex]k[/itex]?
     
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  3. Apr 13, 2012 #2

    Dick

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    Yes, it is. [itex]t^k=e^{log(t) k}[/itex]. If k is pure imaginary and t>0 |t^k|=1.
     
    Last edited: Apr 13, 2012
  4. Apr 14, 2012 #3
    So would you prove it as follows:

    If [itex]x+iy=k\in\mathbb{C}[/itex] and [itex]t,x,y\in\mathbb{R}[/itex] with [itex]t>0[/itex] then [tex]|t^k| = |t^{x+iy}| = |t^x t^{iy}| = |t^x e^{\ln(t)iy}| = |t^x||e^{\ln(t)iy}| = |t^x| = t^x[/tex] so we're just back in the real case where we know [itex]\int_0^a t^x dt[/itex] converges for [itex]x=\text{Re}(k)>-1[/itex].

    It is true that [itex]|t^x| = t^x[/itex] for any real [itex]x[/itex] and positive real [itex]t[/itex] isn't it?
     
    Last edited: Apr 14, 2012
  5. Apr 14, 2012 #4

    Dick

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    Sure. Why would you doubt that?
     
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