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Re: Power Series for LogIntegral

• To: mathgroup at smc.vnet.net
• Subject: [mg45227] Re: [mg45204] Power Series for LogIntegral
• From: Dr Bob <drbob at bigfoot.com>
• Date: Sun, 21 Dec 2003 03:42:19 -0500 (EST)
• References: <200312201056.FAA09567@smc.vnet.net> <E72C1D64-32E3-11D8-A2B6-00039311C1CC@mimuw.edu.pl>
• Reply-to: drbob at bigfoot.com
• Sender: owner-wri-mathgroup at wolfram.com

>> LogIntegral has a brach discontinuity running from -Infinity to 1

I think you mean from -Infinity to 0, as the following gives a power 
series, for instance:

Series[LogIntegral[x], {x, 1/10, 2}]
SeriesData[x, 1/10,
   {ExpIntegralEi[-Log[10]],
    -Log[10]^(-1), 5/Log[10] -
     (5*(1 + Log[10]))/
      Log[10]^2}, 0, 3, 1]

Bobby

On Sat, 20 Dec 2003 20:59:02 +0900, Andrzej Kozlowski <akoz at mimuw.edu.pl> 
wrote:

>
> On 20 Dec 2003, at 19:56, Bobby R. Treat wrote:
>
>> Why doesn't Series return a power series?
>>
>> Series[LogIntegral[x],
>>   {x, 0, 1}]
>>
>> SeriesData[x, 0,
>>   {6/Log[x]^4 + 2/Log[x]^3 +
>>     Log[x]^(-2) + Log[x]^
>>      (-1)}, 1, 2, 1]
>>
>> Bobby
>>
>>
>
> LogIntegral has a brach discontinuity running from -Infinity to 1, so to 
> get a power series you need to expend at a point outside this segment:
>
>
> Normal[Series[LogIntegral[x], {x, 3/2, 2}]]
>
>
> ((-1 + Log[3/2])/(3*Log[3/2]^2) - 2/(3*Log[9/4]))*
>     (x - 3/2)^2 + (2*(x - 3/2))/Log[9/4] +
>    ExpIntegralEi[Log[3/2]]
>
>
> Normal[Series[LogIntegral[x], {x, I, 2}]]
>
>
> (1/Pi - (2*I + Pi)/Pi^2)*(x - I)^2 - (2*I*(x - I))/Pi +
>    ExpIntegralEi[(I*Pi)/2]
>
>
> Andrzej Kozlowski
> Chiba, Japan
> http://www.mimuw.edu.pl/~akoz/
>

• References:
  • Power Series for LogIntegral
    • From: drbob@bigfoot.com (Bobby R. Treat)