# Thread: Integration of unbounded function.

1. ## Integration of unbounded function.

For which values of $p$ the integral $\int^2_1 \frac{x}{(x-1)^p}dx$ converges?

Thank you...

2. What have you tried so far?

3. Nothing actually...

4. Well, I would try computing the integral. Do your normal Calc II stuff and see what happens.

5. ## I solved this!

$\int^2_1\frac{x}{(x-1)^p}dx<\int^2_1\frac{2}{(x-1)^p}dx$

Now, $\int\frac{2}{(x-1)^p}dx=\frac{1}{1-p}\frac{1}{(x-1)^{p-1}}$

By some theorem...
If $p-1<1$ or $p<2$ then the above integral converges.

6. Well, I would have used u = x-1 on the original integral.

7. Originally Posted by Also sprach Zarathustra
$\int^2_1\frac{x}{(x-1)^p}dx<\int^2_1\frac{2}{(x-1)^p}dx$

Now, $\int\frac{2}{(x-1)^p}dx=\frac{1}{1-p}\frac{1}{(x-1)^{p-1}}$

By some theorem...
If $p-1<1$ or $p<2$ then the above integral converges.
Check your work, the answer is $p<1$.

8. Originally Posted by chiph588@
Check your work, the answer is $p<1$.
I disagree!

9. I agree with chiph588. Try u substitution! The result pops right out.

10. Originally Posted by Also sprach Zarathustra
I disagree!
Let $p=1$:

$\int \frac{x}{x-1}dx = x+\ln|x-1|$. What happens when you plug in $x=1$?

11. ## I still don't get it!

...with substetution $x-1=u$.

$\int\frac{x}{(x-1)^p}dx=\int\frac{u+1}{(u)^p}du=\int\frac{1}{u^{p-1}}du+\int\frac{1}{u^p}du=\frac{1}{u^{p-2}}\frac{1}{2-p}+\frac{1}{u^{p-1}}\frac{1}{1-p}$

...then:

$p-2<1$ and $p-1<1$

or:

$p<3$ and $p<2$

Therefore:

$p<2$

12. Originally Posted by Also sprach Zarathustra
...with substetution $x-1=u$.

$\int\frac{x}{(x-1)^p}dx=\int\frac{u+1}{(u)^p}du=\int\frac{1}{u^{p-1}}du+\int\frac{1}{u^p}du=\frac{1}{u^{p-2}}\frac{1}{2-p}+\frac{1}{u^{p-1}}\frac{1}{1-p}$

...then:

$p-2<$ 0 and $p-1<$ 0

or:

$p<3$ and $p<2$

Therefore:

$p<2$
.

13. Why is that???

14. I have got it!!!! silly me!