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MathGroup Archive 2007

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Re: Improper Integral & Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg80621] Re: Improper Integral & Mathematica
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Sun, 26 Aug 2007 23:21:57 -0400 (EDT)
  • References: <faraa2$4k6$1@smc.vnet.net>

On 26    , 10:32, expizzai... at gmail.com wrote:
> Hi all !   :-)
>
> how to solve:
>
> Integrate[(e^x) / (1 + e^(2x)), {x, - Infinity, + Infinity }]
>
> using Mathematica? I'm trying "e^x = t" but won't work...
>
> TNX!

In[7]:=
$Version

Out[7]=
5.2 for Microsoft Windows (June 20, 2005)

I guess you mean E(xponential) instead of e.

If so,

In[8]:=
Integrate[E^x/(1 + E^(2*x)), {x, -Infinity, Plus[Infinity]}]
{N[%], NIntegrate[E^x/(1 + E^(2*x)), {x, -Infinity, Plus[Infinity]}]}
(*check*)

Out[8]=
Pi/2

Out[9]=
{1.5707963267948966, 1.5707963269014895}

With your proposed substitution the result is also the same:

In[18]:=
(E^x/(1 + E^(2*x)))*Dt[x] /. x -> Log[t] /. Dt[t] -> 1
Integrate[%, {t, 0, Infinity}]

Out[18]=
1/(1 + t^2)

Out[19]=
Pi/2

Dimitris

PS

Here is also a nice trick I find it useful many times.
Note however that it is not pannacea, and you must
be very careful when you use it.

Say you indeed mean e and not E(xponential).

Then the integral stays unevaluated.

In[66]:=
Integrate[e^x/(1 + e^(2*x)), {x, -Infinity, Plus[Infinity]}]

Out[66]=
Integrate[e^x/(1 + e^(2*x)), {x, -Infinity, Infinity}]

However,

In[67]:=
Integrate[EulerGamma^x/(1 + EulerGamma^(2*x)), {x, -Infinity,
Plus[Infinity]}]
{N[%], NIntegrate[EulerGamma^x/(1 + EulerGamma^(2*x)), {x, -Infinity,
Plus[Infinity]}]}
res = %% /. EulerGamma -> e

Out[67]=
-(Pi/(2*Log[EulerGamma]))

Out[68]=
{2.858387543326428, 2.8583875434363413}

Out[69]=
-(Pi/(2*Log[e]))

So using this nice trick (replace e with EulerGamma, evaluate the
integral
and return to the original parameter) we have -(Pi/(2*Log[e])) as the
result.
Which is valid at least for 0<e<1 as the following shows.

In[72]:=
res/.e\[Rule]#&/@Range[0.1,0.9,0.1]
NIntegrate[(#^x)/(1+#^(2x)),{x,-Infinity,+Infinity}]&/
@Range[0.1,0.9,0.1]

Out[72]=
{0.682188,0.975991,1.30468,1.7143,2.26618,3.07501,4.404,7.0394,14.9088}

Out[73]=
{0.682188,0.975991,1.30468,1.7143,2.26618,3.07501,4.404,7.0394,14.9088}




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