1. ## improper integral

I'm given an integral whose bounds are from e to infinity of dx/(x*(lnx)^a) and I have to find for which values of a it converges or diverges. How do you deal with the log in there? the only test I know is a comparison test but it didn't help when I tried it

2. Do you know for what values of a $\displaystyle \displaystyle \int_1^\infty {\frac{{du}}{{u^a }}}$ converges?

3. http://orion.math.iastate.edu/vika/cal3_files/Lec22.pdf

That should help - look at the last page. Essentially, if a>1 the infinite series is convergent. So for a<a1 the infinite series is divergent. The proof is within the link I provided.

4. Originally Posted by BariMutation
http://orion.math.iastate.edu/vika/cal3_files/Lec22.pdf

That should help - look at the last page. Essentially, if a>1 the infinite series is convergent. So for a<a1 the infinite series is divergent. The proof is within the link I provided.
What is a primitive of $\displaystyle \displaystyle \frac{1}{x\ \ln^{a} x}$ ?... consulting the book You find...

$\displaystyle \displaystyle \int \frac{dx}{x\ \ln^{a} x} = \left\{\begin{array}{ll}\frac {\ln ^{1-a} x}{1-a},\,\, a\ne 1\\{}\\ \ln |\ln x| ,\,\, a=1\end{array}\right.$ (1)

Now if You valuate the limit $\displaystyle x \rightarrow \infty$ of (1), You 'discover' that the definite integral converges for $\displaystyle a>1$ and diverges for $\displaystyle a \le 1$...

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$

I have another one now :S
integral from 0 to infinity of the function (t^(a-1))*(e^(-t)), again find which values of a it converges/diverges

6. and another still. I'm having a lot of trouble with these ]:
integral from 0 to infinity of (x^(a-1))/(1+x)
help would be greatly appreciated

I have another one now :S
integral from 0 to infinity of the function (t^(a-1))*(e^(-t)), again find which values of a it converges/diverges
The question is a little controverisial... one way to 'attack' the problem is the following: let suppose to define the function $\displaystyle \Gamma(a)$ as...

$\displaystyle \displaystyle \Gamma(a)= \int_{0}^{\infty} t^{a-1}\ e^{-t}\ dt$ (1)

Now integrating by parts (1) You obtain...

$\displaystyle \displaystyle \int_{0}^{\infty} t^{a-1}\ e^{-t}\ dt = \frac{1}{a}\ \int_{0}^{\infty} t^{a}\ e^{-t}\ dt$ (2)

... or equivalently...

$\displaystyle \displaystyle \Gamma(a)= \frac{\Gamma(a+1)}{a}$ (3)

Observing (3) You conclude that...

$\displaystyle \displaystyle \lim_{a \rightarrow 0+} \Gamma(a)= +\infty$ (4)

... so that may seem obvious that the integral converges for $\displaystyle a>0$ and diverges for $\displaystyle a \le 0$... therefore for someone [a little suprisingly!...] that's not obvious at all and that's a long story ...

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$

and another still. I'm having a lot of trouble with these ]:
integral from 0 to infinity of (x^(a-1))/(1+x)
help would be greatly appreciated
Integrating by parts You obtain...

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

... and observing the first tem of (1) You can conclude that the integral converges for $\displaystyle 0<a<1$...

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$