# Improper Integrals

• Jan 18th 2009, 04:34 PM
zangestu888
Improper Integrals
The integral of

e^x / e^2x + 1 from upper limit:infinite and lower limit:0

So for I have the antderirive as:

this after changin the integration limits to a limit

limit t--> (arctan(e^x)) from b=infinte and a=0

and also this one

the integral of

ln(x)/x^2 from upper limit=b=infinte and lower limit=a=1 not sure how to start this one at all!
• Jan 18th 2009, 05:04 PM
o_O
(1) Recall that: $\lim_{x \to \infty} \arctan x = \frac{\pi}{2}$

So: $\lim_{t \to \infty} \arctan x \Big|_{0}^{t} \ = \ \lim_{t \to \infty} \left( \arctan t - \arctan 0\right) \ = \ \cdots$

(2) To integrate, use integration by parts: $\begin{array}{lll} u = \ln x & \quad & dv = \frac{1}{x^2} \ dx \\ du = \frac{1}{x}dx & \quad & v = -\frac{1}{x} \end{array}$

See if you can carry on from there.
• Jan 18th 2009, 05:07 PM
Krizalid
$\frac{{{e}^{x}}}{{{e}^{2x}}+1}\le \frac{{{e}^{x}}}{{{e}^{2x}}}={{e}^{-x}},$ hence $\int_{0}^{\infty }{\frac{{{e}^{x}}}{{{e}^{2x}}+1}\,dx}$ converges.
• Jan 18th 2009, 05:14 PM
zangestu888
..hmm
Okay so am guessing my first part was correct, now I have to evaulate the integral as it approces infinite, I have one more question this is my weakness. am not sure how to exactly evaulte limits so for in my case:

lim
t-->arctan(e^x)=limt --> infinite (arctan(e^t) - arctan(o))
=lim
t->infinte (arctan(e^t) -0))

=lim t-->infinte (arctan(e^t))=pi/2 ?? i think am write :D lol that ws great help am gonna work on the other one now ! thanks!BANKAI
• Jan 18th 2009, 05:17 PM
o_O
Yep, as $t \rightarrow \infty$ then $e^t \rightarrow \infty$ as well. So $\lim_{t \to \infty} \arctan e^t = \frac{\pi}{2}$

Good job, Kurosaki Ichigo.
• Jan 18th 2009, 09:16 PM
chiph588@
You forgot that it's $\tan^{-1} (e^0)$ not $\tan^{-1} (0)$

$\lim_{t \to \infty} \tan^{-1} (t) - \tan^{-1} (1) = \frac{\pi}{2}-\frac{\pi}{4} = \frac{\pi}{4}$