Thread: Improper Integrals (and one that isn't!)

1. Improper Integrals (and one that isn't!)

1) $\displaystyle \int_{0}^{1}\frac{1}{x*ln{x}}dx$

2) $\displaystyle \int_{0}^{\infty}\frac{x}{e^{x}}dx$

3) $\displaystyle \int_{0}^{\infty}\frac{1}{x*(ln{x})^{2}}dx$

These beasts are the only thing I've yet to conquer on my latest AP Calc BC take home test. I've thought about integration by parts and even simple u-substitution but I'm drawing a blank here. I'm somewhat sure I need to substitute in a value for infinity and take the limit as that value goes to infinity for the impropers, but I'm not really sure.

Any help is appreciated. :>

2. Let's do the second one.

$\displaystyle \int_{0}^{\infty}\frac{x}{e^{x}}dx$

$\displaystyle \int_{0}^{L}\frac{x}{e^{x}}dx$

$\displaystyle =\frac{-L}{e^{L}}-\frac{1}{e^{L}}+1$

$\displaystyle \lim_{L\rightarrow{L}}\left[\frac{-L}{e^{L}}-\frac{1}{e^{L}}+1\right]=\boxed{1}$

As you should be able to see, as L is unbounded, we are left with 1

3. Thanks, I thought it was something along those lines. I also edited my original post to make it more read-friendly.

Also, would you mind elaborating more on how you found the anti-derivative of $\displaystyle \int_{0}^{L}\frac{x}{e^{x}}dx$? Sorry, I just don't see it. :<

4. Hello, zerot!

$\displaystyle 1)\;\;\int_{0}^{1}\frac{1}{x\ln x}dx$
We have: .$\displaystyle \int\frac{1}{\ln x}\cdot\frac{dx}{x}$

Let: .$\displaystyle u \:=\:\ln x\quad\Rightarrow\quad du \:=\:\frac{dx}{x}$

Substitute: .$\displaystyle \int\frac{1}{u}\,du$ . . . . Got it?

$\displaystyle 2)\;\;\int_{0}^{\infty}\frac{x}{e^x}dx$
This one requires by-parts . . .

$\displaystyle \begin{array}{ccccccc} u & = & x & \quad & dv & = & e^{-x}dx \\ du & = & dx & \quad & v & = & -e^{-x} \end{array}$

We have: .$\displaystyle -xe^{-x} + \int e^{-x}dx$ . . . . Carry on!

$\displaystyle 3)\;\;\int_{0}^{\infty}\frac{1}{x(\ln x)^{2}}dx$
This is the same the first problem . . .

We have: .$\displaystyle \int\frac{1}{(\ln x)^2}\,\frac{dx}{x}$

Let: .$\displaystyle u \:=\:\ln x\quad\Rightarrow\quad du \:=\:\frac{dx}{x}$

Substitute: .$\displaystyle \int\frac{1}{u^2}\,du \;=\;\int u^{-2}du$ . . . . Okay?

5. The problem is that I cannot evaluate the definite integral. I did the first and third problems the way you also did, but that leaves me finding ln|0| and ln|ln|0|| and that's what's hanging me up.