       Re: Integrate with PrincipalValue->True

• To: mathgroup at smc.vnet.net
• Subject: [mg80003] Re: Integrate with PrincipalValue->True
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Fri, 10 Aug 2007 01:55:24 -0400 (EDT)
• Organization: The Open University, Milton Keynes, UK
• References: <f9em11\$k0v\$1@smc.vnet.net>

```chuck009 wrote:
> Can someone explain to me how Integrate calculates the following improper integral?
>
> In:= \$Version
> Integrate[Sqrt[x]/(1 - x^2), {x, 0, a},
>    PrincipalValue -> True, Assumptions ->
>      a > 1]
>
> Out= "6.0 for Microsoft Windows (32-bit) (June 19, 2007)"
>
> Out= ArcCoth[Sqrt[a]] - ArcTan[Sqrt[a]]

Charles,

algorithms or about Cauchy principal value integral.

In your case, setting the option PrincipalValue to true is conceptually
equivalent to computing the improper integral as follows.

In:= \$Version
f[x_] = Sqrt[x]/(1 - x^2);
pv = Limit[Integrate[f[x], {x, 0, 1 - epsilon},
Assumptions -> epsilon > 0] +
Integrate[f[x], {x, 1 + epsilon, a},
Assumptions -> epsilon > 0 && a > 1], epsilon -> 0,
Assumptions -> a > 1]

Out= "6.0 for Microsoft Windows (32-bit) (June 19, 2007)"

Out= -(\[Pi]/2) + ArcTan[1/Sqrt[a]] + ArcTanh[1/Sqrt[a]]

(You can check numerically or symbolically that this is equivalent to