Re: Why does this lead to an answer with complex numbers?

• To: mathgroup at smc.vnet.net
• Subject: [mg71497] Re: Why does this lead to an answer with complex numbers?
• From: "David W. Cantrell" <DWCantrell at sigmaxi.net>
• Date: Mon, 20 Nov 2006 18:12:00 -0500 (EST)
• Organization: NewsReader.Com Subscriber
• References: <ejosmm\$n3k\$1@smc.vnet.net> <ejrmr9\$97b\$1@smc.vnet.net> <ejs3fv\$9ii\$1@smc.vnet.net>

```Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com> wrote:
> aaronfude at gmail.com wrote:
> > Hi,
> >
> > Thanks for all the answers. They were all very useful, even though I
> > have done my best to confuse everyone by leaving a beta in there which
> > had nothing to do with the problem.
> >
> > So I understand that the answer may be complex and the complex part is
> > constant which is in a certain sense valid for a indefinite integral.
> > But I very much need a real answer and I still can't quite extract.
> > Consider the following:
> >
> > \!\(\(\(\ \)\(Assuming[x > 0 && A > 0 && B > 0 && \ B < 1, \
> >     FullSimplify[Integrate[Log[\@\(A^2 + x\^2\) - B*x\ ], \ x]]]\)\)\)
> >
> > The answer that I get is correct, but not very useful since it is
> > appears complex and I could find a way to determine the real part. Do
> > you have any suggestions?

For mine, see the end of this post.

> Hi Aaron,
>
> What version of Mathematica do you use? With Mathematica 5.2 I do not
> get any complex numbers for your integral.

Yes, you do. The mere fact that /I/ does not appear in the expression does
not mean that it is necessarily real. While there may be some special
values of A, B, and x which cause Out[1] to be real, it is generally
complex.

> In[1]:=
> Assuming[x > 0 && A > 0 && B > 0 && B < 1,
>    FullSimplify[Integrate[Log[Sqrt[A^2 + x^2] - B*x], x]]]
>
> Out[1]=
>         1                      2                           2    2
> --------------- (2 Sqrt[-1 + B ] x (-1 + Log[-B x + Sqrt[A  + x ]]) +
>               2
> 2 Sqrt[-1 + B ]
>
>                              2
>                   Sqrt[-1 + B ] x
>      A (2 ArcTanh[---------------] - 4 Log[A] - 4 Log[B] +
>                          A
>
>                                                2                2    2
>               4 (-1 + B) (1 + B) (A Sqrt[-1 + B ] - x + B Sqrt[A  + x ])
>
> Log[-(----------------------------------------------------------)] +
>                                          2           2
>                             A Sqrt[-1 + B ] + (-1 + B ) x
>
>                    2                2                2    2
>             (-1 + B ) (A Sqrt[-1 + B ] + x + B Sqrt[A  + x ])
>         Log[-------------------------------------------------]))
>                                     2         2
>                        A Sqrt[-1 + B ] + x - B  x
>
> In[2]:=
> FreeQ[%, I]
>
> Out[2]=
> True
[snip]

My suggestion is that you get an antiderivative using some work done by
hand. My result is

x*(Log[Sqrt[A^2 + x^2] - B*x] - 1) +
A/Sqrt[1 - B^2]*(ArcCot[A*B/Sqrt[(1 - B^2)*(A^2 + x^2)]] +
ArcTan[Sqrt[1 - B^2]*x/A])

the correctness of which is easily checked using Mathematica:

In[1]:= FullSimplify[D[x*(Log[Sqrt[A^2 + x^2] - B*x] - 1) +
A/Sqrt[1 - B^2]*(ArcCot[A*B/Sqrt[(1 - B^2)*(A^2 + x^2)]] +
ArcTan[Sqrt[1 - B^2]*x/A]), x], x > 0 && A > 0 && 0 < B < 1]

Out[1]= Log[(-B)*x + Sqrt[A^2 + x^2]]

Note that my antiderivative is not only real, but all its subexpressions
are real. Can anyone get Mathematica to find such an antiderivative
directly?

David

```

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