Integration of unbounded function.

**Also sprach Zarathustra** · June 24th 2010, 07:04 AM

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

Thank you...

**Ackbeet** · June 24th 2010, 07:08 AM

What have you tried so far?

**Also sprach Zarathustra** · June 24th 2010, 07:14 AM

Nothing actually...

**Ackbeet** · June 24th 2010, 07:15 AM

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

**Also sprach Zarathustra** · June 24th 2010, 07:35 AM

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

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.

**Ackbeet** · June 24th 2010, 07:36 AM

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

**chiph588@** · June 24th 2010, 07:51 AM

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 $p<1$ .

**Also sprach Zarathustra** · June 24th 2010, 07:58 AM

Originally Posted by chiph588@

Check your work, the answer is $p<1$ .

I disagree!

**Ackbeet** · June 24th 2010, 08:02 AM

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

**chiph588@** · June 24th 2010, 08:12 AM

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 $x=1$ ?

**Also sprach Zarathustra** · June 24th 2010, 08:56 AM

...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:

$p<2$

**chiph588@** · June 24th 2010, 08:58 AM

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:

$p<2$

.

**Also sprach Zarathustra** · June 24th 2010, 09:05 AM

Why is that???

**Also sprach Zarathustra** · June 24th 2010, 09:12 AM

I have got it!!!! silly me!

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Integration of unbounded function.

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