Re: Piecewise and Integral

• To: mathgroup at smc.vnet.net
• Subject: [mg75256] Re: Piecewise and Integral
• From: Bhuvanesh <lalu_bhatt at yahoo.com>
• Date: Sun, 22 Apr 2007 05:11:10 -0400 (EDT)

```First, conditional assignments such as

f[x_]:=f[x-T]/;x >T/2

are not used by Integrate in any way. Integrate uses whatever you get by evaluating f[x], that is,

Piecewise[{{0, Inequality[-1, LessEqual, x, Less, -1/2]}, {10, -1/2 <= x <= 1/2}}]

When \[Lambda]==T/2==1, the integration limits are from 0 to 2, so we get:

Integrate[Piecewise[{{0, Inequality[-1, LessEqual, x, Less, -1/2]}, {10, -1/2 <= x <= 1/2}}], {x,0,2}]

which is Integrate[10, {x,0,1/2}] == 5.

In order to get what you want, you have to include the periodicity information in the initial function definition, e.g. something like

f[x_] = 10*UnitStep[Sin[Pi*x + 3/2]]

Bhuvanesh,
Wolfram Research

```

• Prev by Date: question
• Next by Date: strange behavior of Integrate
• Previous by thread: Re: Piecewise and Integral
• Next by thread: Re: Piecewise and Integral