# The dummy variable

• Dec 22nd 2009, 12:40 PM
rainer
The dummy variable

$\int_a^b{g(x)} dt = b*g(x) - a*g(x)$

When I apply the rule to the following equation--

$\frac{\pi}{2x}=\int_{0}^{\frac{\pi}{2}}\frac{\pi\l eft(1+\tanh\left(\frac{-\ln{\sin\theta}}{2} \right ) \right )}{4x} d\theta$

I get:

$\frac{\pi}{2}\frac{\pi\left(1+\tanh\left(\frac{-\ln{\sin\frac{\pi}{2}}}{2} \right ) \right )}{4x}=\frac{\pi^2}{8x}\neq\frac{\pi}{2x}$

which doesn't make sense. What am I missing?

2) How does the rule work for definite integrals whose intervals are infinite, such as in the gamma function?

Pardon my ignorance.
• Dec 22nd 2009, 01:28 PM
thelostchild
The rule you posted only works for when the variable that your integrating isn't in the function itself it can be shown as

$
\int_{a}^{b} g(x) dt = g(x) \int_{a}^{b} dt = g(x) \times (a-b)
$

so it would be only applicable if you can do that first step.

In the example you gave you cannot because your integrating with respect to theta so cannot factor out that function of theta
• Dec 23rd 2009, 09:17 AM
rainer
Quote:

Originally Posted by thelostchild
The rule you posted only works for when the variable that your integrating isn't in the function itself it can be shown as

$
\int_{a}^{b} g(x) dt = g(x) \int_{a}^{b} dt = g(x) \times (a-b)
$

so it would be only applicable if you can do that first step.

In the example you gave you cannot because your integrating with respect to theta so cannot factor out that function of theta

Thanks. But I'm still not entirely clear on how you can integrate with respect to a variable that is not even in the equation. I would be much obliged if you could give a simple example.
• Dec 24th 2009, 02:01 AM
Moo
Quote:

Originally Posted by rainer
Thanks. But I'm still not entirely clear on how you can integrate with respect to a variable that is not even in the equation. I would be much obliged if you could give a simple example.

If you want something you know :

An antiderivative of 1 is t.

So $\int_a^b ~dt=\int_a^b 1 ~dt=\left. t\right|_a^b=b-a$

Satisfied ?