Re: Integrate + Conjugate = Indeterminate

• To: mathgroup at smc.vnet.net
• Subject: [mg120661] Re: Integrate + Conjugate = Indeterminate
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Wed, 3 Aug 2011 07:04:49 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <j0tru0\$d4m\$1@smc.vnet.net> <201108021112.HAA29358@smc.vnet.net>

```On 08/02/2011 06:12 AM, Sebastian Hofer wrote:
> I know that my question was originally posed in a very vague way, which was partly due to me having no idea where my problems came from. After more analysis of the involved expressions, I managed to get them into a form which can be evaluated numerically without getting Indeterminate values. What I essentially did was to use a combination of iterated Expands and replacement rules like
>
> Expand[#] /. Power[E, x_] :>  Power[E, FullSimplify@Expand@x]
>
> (the expressions involve several similar exponentials which are multiplied and added...). Also, as the whole expression is the result of an integral, I separated the sub-expression which needed to be integrated before the integration and put everything back together afterwards.
>
> Also, I was bitten by (let's call it) a very peculiar "feature" of Mathematica's handling of Conjugate:
>
> In[25]:= Integrate[
>   Exp[(Sqrt[a + I b] - Conjugate[Sqrt[a - I b]]) x], {x, x0, x1},
>   Assumptions ->  a>  0&&  b>  0]
>
> Out[25]= (-E^(
>    x0 (Sqrt[a + I b] - Conjugate[Sqrt[a - I b]])) + E^(
>   x1 (Sqrt[a + I b] - Conjugate[Sqrt[a - I b]])))/(Sqrt[a + I b] -
>   Conjugate[Sqrt[a - I b]])
>
> In[23]:= Out[18] /. {a ->  1, b ->  2} // N
>
> During evaluation of (Local 2) In[23]:= Power::infy: Infinite expression 1/(0. +0. I)
>   encountered.>>
>
> During evaluation of (Local 2) In[23]:= Power::infy: Infinite expression 1/(0. +0. I)^1.
>   encountered.>>
>
> During evaluation of (Local 2) In[23]:= Infinity::indet: Indeterminate expression (0. +0. I) ComplexInfinity
>   encountered.>>
>
> Out[23]= Indeterminate
>
> Of course this holds for all a,b. Is there any logical explanation for this? Or better: Is there a way to get around it? This problem is probably known to many of you, but it was new to me and I would think it is a bug...
>
> Sebastian

The reason behind it is that Mathematica fails to recognize that (Sqrt[a
+ I b] - Conjugate[Sqrt[a - I b]]) is zero when a,b>0.

Daniel Lichtblau
Wolfram Research

```

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