# Integral of sech(x) between infinity and minus infinity

• May 8th 2011, 03:44 AM
incblue
Integral of sech(x) between infinity and minus infinity
As the title says, I am trying to find the definite integral of sech(x) with limits of infinity and minus infinity.

I always get 2pi as my solution, but all calculators and integration applets give pi as the answer.

Below is my method so far:
http://img853.imageshack.us/img853/3149/capturehkr.png

Could someone please tell me where I have gone wrong?

EDIT: there is an error in line 6 of my method, that extra u is just a typo.
• May 8th 2011, 03:47 AM
Prove It
You didn't change your terminals when you substituted u. Also, your integrand should be 1/(u^2 + 1), not u/(u^2 + 1), but since you have integrated this correctly I suspect that is just a typo.
• May 8th 2011, 05:10 AM
incblue
Quote:

You didn't change your terminals when you substituted u.
I did, but e^infinity is just infinity.

Quote:

Also, your integrand should be 1/(u^2 + 1), not u/(u^2 + 1), but since you have integrated this correctly I suspect that is just a typo.
Yes, that was a typo. I had mentioned it in an edit prior to your post.

Anyway, I still have the issue of 2pi versus pi.
• May 8th 2011, 05:13 AM
Prove It
Quote:

Originally Posted by incblue
I did, but e^infinity is just infinity.

But $\displaystyle \displaystyle \lim_{x \to -\infty}e^x = 0$, not $\displaystyle \displaystyle -\infty$...