# Integrating hyperbolic function to infinity

• May 9th 2009, 12:16 PM
james
Integrating hyperbolic function to infinity
I need to find
$\displaystyle \int_0^{\infty}1-\tanh\left(\frac{x}{a}\right) dx$
which I get to be
$\displaystyle x-a \ln\left(\cosh \left(\frac{x}{a}\right)\right)$
But I'm not sure how to use the limit at infinity for this. I'm also pretty sure I'm not meant to be getting zero at the answer but something depending on a.
Any help would be greatly appreciated!
• May 9th 2009, 12:20 PM
Banned for attempted hacking
Quote:

Originally Posted by james
I need to find
$\displaystyle \int_0^{\infty}1-\tanh\left(\frac{x}{a}\right) dx$
which I get to be
$\displaystyle x-a \ln\left(\cosh \left(\frac{x}{a}\right)\right)$
But I'm not sure how to use the limit at infinity for this. I'm also pretty sure I'm not meant to be getting zero at the answer but something depending on a.
Any help would be greatly appreciated!

Remember that when it's an improper integral you have to take the limit as lets say b---> infinity.

An easy way to do this is to use the fuction of cosh = $\displaystyle \frac{e^{x}+e^{-x}}{2}$

You may have to use l'hospitals for the limit since you might end up with a form infinity - infinity.
• May 9th 2009, 12:46 PM
james
Thanks for the quick reply, think I've got somethingwhich looks alright. What do you think?

$\displaystyle x-a \ln\left(\cosh \left(\frac{x}{a}\right)\right)$
over b and zero gives
$\displaystyle b-a \ln\left(\cosh \left(\frac{b}{a}\right)\right) - 0 + a \ln\left(\cosh \left(0\right)\right)$
$\displaystyle = b - a \ln \left(\frac{e^{b/a}+e^{-b/a}}{2} \right)$
$\displaystyle = b - a \ln \left(e^{b/a}+e^{-b/a} \right) + a \ln 2$
then as b -> infinity the negative exponential vanishes so
$\displaystyle = b - a \ln \left(e^{b/a}\right) + a \ln 2$
$\displaystyle = b - a (b/a) + a \ln 2$
$\displaystyle = a \ln 2$
• May 9th 2009, 12:50 PM
Banned for attempted hacking
Quote:

Originally Posted by james
Thanks for the quick reply, think I've got somethingwhich looks alright. What do you think?

$\displaystyle x-a \ln\left(\cosh \left(\frac{x}{a}\right)\right)$
over b and zero gives
$\displaystyle b-a \ln\left(\cosh \left(\frac{b}{a}\right)\right) - 0 + a \ln\left(\cosh \left(0\right)\right)$
$\displaystyle = b - a \ln \left(\frac{e^{b/a}+e^{-b/a}}{2} \right)$
$\displaystyle = b - a \ln \left(e^{b/a}+e^{-b/a} \right) + a \ln 2$
then as b -> infinity the negative exponential vanishes so
$\displaystyle = b - a \ln \left(e^{b/a}\right) + a \ln 2$
$\displaystyle = b - a (b/a) + a \ln 2$
$\displaystyle = a \ln 2$

That looks right and funny at the same time. :P
• May 9th 2009, 12:55 PM
james
Well as long as it's right I don't mind people laughing at my work...

Thanks a lot :)
• May 9th 2009, 01:04 PM
Banned for attempted hacking
I said funny because it becomes b -b = 0 which is equilalent to
infinity - infinity which is supposed to be undefined.