improper intergrals

**momo55** · June 10th 2010, 08:35 PM

i am trying to see if i am right on this answer i was asked to see if
the intergral of e^x/(e^x-1) from -1 to 1 is convergent or divergent i
think it is divergent because 0 lies between the the limits given and it would
make the denominator equal to 0 which is undefined

p.s sry i had to write the problem like this because i do not know how to write
the integral in the message box like others in this forum
thank you for the help

**simplependulum** · June 10th 2010, 09:09 PM

Originally Posted by momo55

i am trying to see if i am right on this answer i was asked to see if
the intergral of e^x/(e^x-1) from -1 to 1 is convergent or divergent i
think it is divergent because 0 lies between the the limits given and it would
make the denominator equal to 0 which is undefined

p.s sry i had to write the problem like this because i do not know how to write
the integral in the message box like others in this forum
thank you for the help

I am not keen on such improper integral but can we do it in this way ?

$\int_{-1}^1 \frac{e^{x}}{e^x - 1}~dx$

$= \lim_{a\to 0^+} \left( \int_a^1 \frac{e^{x}}{e^x - 1}~dx + \int_{-1}^{-a} \frac{e^{x}}{e^x - 1}~dx \right)$

Sub . $x = -t$ on the second integral we have

$= \lim_{a\to 0^+} \left( \int_a^1 \frac{e^{x}}{e^x - 1}~dx + \int_a^1 \frac{e^{-x}}{e^{-x} - 1}~dx \right)$

$= \lim_{a\to 0^+} \left( \int_a^1 \frac{e^{x}}{e^x - 1}~dx + \int_a^1 \frac{1}{1 - e^x}~dx \right)$

$= \lim_{a\to 0^+} \int_a^1 \left( \frac{e^{x}}{e^x - 1} - \frac{1}{e^x - 1 } \right) ~dx$

$= \lim_{a\to 0^+} \int_a^1 ~dx = \lim_{a\to 0^+} [x]_a^1 = \lim_{a\to 0^+} (1-a) = 1$

**chisigma** · June 10th 2010, 10:51 PM

Originally Posted by simplependulum

I am not keen on such improper integral but can we do it in this way ?

$\int_{-1}^1 \frac{e^{x}}{e^x - 1}~dx$

$= \lim_{a\to 0^+} \left( \int_a^1 \frac{e^{x}}{e^x - 1}~dx + \int_{-1}^{-a} \frac{e^{x}}{e^x - 1}~dx \right)$

...

The question is a little controversial... what simplependulum has computed is the so called 'principal value' of the integral...

$\int_{-1}^1 \frac{e^{x}}{e^x - 1}\cdot dx$ (1)

For more information see...

Cauchy Principal Value -- from Wolfram MathWorld

In general if we don't specify that, is...

$\int_{-1}^1 \frac{e^{x}}{e^x - 1}\cdot dx = \lim_{a \rightarrow 0-} \int_{-1}^{a} \frac{e^{x}}{e^x - 1}\cdot dx + \lim_{b \rightarrow 0 +} \int_{b}^{1} \frac{e^{x}}{e^x - 1}\cdot dx$ (2)

... and in this case both the limits in (2) doesn't exist...

Kind regards

$\chi$ $\sigma$

**momo55** · June 11th 2010, 06:43 AM

is it because when you put 0 in the limit the bottom will turn to 0 which will make it undefined? so it diverges ??

**chisigma** · June 11th 2010, 07:12 AM

Originally Posted by momo55

is it because when you put 0 in the limit the bottom will turn to 0 which will make it undefined? so it diverges ??

Yes, it is!...

If You intend to compute the 'principal value' is...

$S= PV \int_{-1}^{1} f(x)\cdot dx = \int_{0}^{1} [f(x) + f(-x)]\cdot dx$ (1)

... and here is...

$f(x) + f(-x) = \frac{e^{x}}{e^{x}-1} + \frac{e^{-x}}{e^{-x} -1} = \frac{e^{x}}{e^{x}-1} - \frac{1}{e^{x}-1} = 1$ (2)

... so that is...

$S= PV \int_{-1}^{1} \frac{e^{x}}{e^{x}-1}\cdot dx = 1$ (3)

Kind regards

$\chi$ $\sigma$

**p0oint** · June 11th 2010, 08:44 AM

I think that there is much easier way out:

$<br /> \int_{-1}^1 \frac{e^{x}}{e^x - 1}~dx<br />$

First compute the antiderivative.

Make substitution $t=e^{x}-1$ , $dt=e^{x}dx$

$<br /> \int \frac{dt}{t}dx = ln|t|=ln|e^{x}-1|<br />$

Because you should find the integral on interval [-1,1] and $ln|e^{x}-1|$ is only defined on $e^{x} \neq 1$ or $x \neq 0$ , clearly the integral do not converge on [-1,1].

Regards.

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