Re: Integration anomaly?

• To: mathgroup at smc.vnet.net
• Subject: [mg127016] Re: Integration anomaly?
• From: Andrzej Kozlowski <akozlowski at gmail.com>
• Date: Sun, 24 Jun 2012 04:25:06 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201206230816.EAA00864@smc.vnet.net>

```This is indeed a bug and Mathematica knows that ;-). Try:

Integrate[Log[1/t^n + 1], {t, 0, Infinity},
Assumptions -> n > 0]

Pi*Csc[Pi/n]

Integrate[Log[1/t^n + 1], {t, 0, Infinity},
Assumptions -> n < 0]

During evaluation of In[63]:= Integrate::idiv:Integral of log(t^-n+1) does not converge on {0,\[Infinity]}. >>

Integrate[Log[t^(-n) + 1], {t, 0, Infinity},
Assumptions -> n < 0]

Andrzej Kozlowski

On 23 Jun 2012, at 10:16, Szumiloski, John wrote:

> Greetings.
>
> I recently was playing around with the function Log[1 + 1 / ( t^n ) ], and exploring positive values of n.  (I have no interest in nonpositive or complex n)   In particular, I wanted to look at its integral, so I did this: (v8.0.4, Windows XP)
>
>
>              Integrate[ Log[1+1/(t^n)], { t, 0, Infinity } ]
>
>
> which gave:
>
>             ConditionalExpression[ -(Pi Csc[Pi/n]), Re[n]<0 ]
>
>
> Now I am no analysis expert, but it seems pretty clear to me that the integral diverges for negative (real) n.
>
> After doing some further specific symbolic and numeric integrations, I suspect the correct value of this integral is actually
>
>
>             ConditionalExpression[ Pi Csc[Pi/n], Re[n]>1 ]
>
>
> although I cannot prove it.
>
> But in any case, the answer Mathematica gave seems clearly wrong.  Can anyone reproduce this (and thus, should someone at WRI look into this), or is my system (or my brain) quirking out?
>
> Thanks,
> John  Szumiloski,  Ph.D.
>
> Senior Biometrician
> Biometrics Research
> WP53B-120
> Merck Research Laboratories
> P.O. Box 0004
> West Point, PA 19486-0004
> USA
> (215) 652-7346 (PH)
> (215) 993-1835 (FAX)
> lowercasefirstname<dot>lowercaselastname<at>merck<dot>com
> ___________________________________________________
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>
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