# Convergent/Divergent Integrals

• February 3rd 2008, 09:53 AM
TrevorP
Convergent/Divergent Integrals
I'm asked to determine if the following integral is convergent/divergent:

$\int_0^1 \ln(x) dx$

So I'm just wondering if I'm on the right track with what I'm doing:

$\int_0^1 \ln(x) dx = \lim_{t \rightarrow 0} \int_t^1 \ln(x) dx$

As $t \rightarrow 0 \Longrightarrow \ln(x) \rightarrow \infty$

So can I then just say this??:

$\int_0^1 \ln(x) dx = \infty \ \ \therefore \ \ \boxed{\mbox{Divergent}}$
• February 3rd 2008, 10:55 AM
mr fantastic
Quote:

Originally Posted by TrevorP
I'm asked to determine if the following integral is convergent/divergent:

$\int_0^1 \ln(x) dx$

So I'm just wondering if I'm on the right track with what I'm doing:

$\int_0^1 \ln(x) dx = \lim_{t \rightarrow 0} \int_t^1 \ln(x) dx$

As $t \rightarrow 0 \Longrightarrow \ln(x) \rightarrow \infty$

So can I then just say this??:

$\int_0^1 \ln(x) \, dx = \infty \ \ \therefore \ \ \boxed{\mbox{Divergent}}$

You might just want to calculate $\int_t^1 \ln(x) dx$ and then look at the limit as $t \rightarrow 0^+$. The limit you need to carefully consider is $\lim_{t \rightarrow 0^+} (t \ln t)$, which is an indeterminant form ......

Note: $\lim_{t \rightarrow 0^+} t \ln t = \lim_{t \rightarrow 0^+} \frac{\ln t}{\frac{1}{t}} ......$ Are you familiar with l'Hopital's Rule?
• February 3rd 2008, 11:00 AM
Krizalid
It converges.

You may perform the straightforward integration by parts way, but I'll perform three ways to tackle this.

Solution #1:

$\int_0^1 {\ln x\,dx} .$ Substitute $u=-\ln x,$

$\int_0^1 {\ln x\,dx} = - \int_0^\infty {ue^{ - u} \,du} = - \Gamma (1) = - 1.$

Solution #2:

$\ln x = \int_1^x {\frac{1}
{u}\,du} .$
Construct a double integral to reverse integration order, the rest follows easily.

Solution #3:

By the geometric series

$\ln (1+x) = \int_0^x {\frac{1}
{{1 + u}}\,du} = \sum\limits_{k = 0}^\infty {( - 1)^k \left( {\int_0^x {u^k \,du} } \right)} = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{k + 1} }}
{{k + 1}}} .$

This yields $\ln x = \sum\limits_{k = 1}^\infty {\frac{{( - 1)^{k + 1} (x - 1)^k }}
{k}} .$
Hence

$\int_0^1 {\ln x \,dx }= \sum\limits_{k = 1}^\infty {\frac{{( - 1)^{k + 1} }}
{k}\left( {\int_0^1 {(x - 1)^k \,dx} } \right)} = - \sum\limits_{k = 1}^\infty {\frac{1}
{{k(k + 1)}}} = - 1.$

The last sum is a telescoping series.
• February 3rd 2008, 02:28 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
You might just want to calculate $\int_t^1 \ln(x) dx$ and then look at the limit as $t \rightarrow 0^+$. The limit you need to carefully consider is $\lim_{t \rightarrow 0^+} (t \ln t)$, which is an indeterminant form ......

Note: $\lim_{t \rightarrow 0^+} t \ln t = \lim_{t \rightarrow 0^+} \frac{\ln t}{\frac{1}{t}} ......$ Are you familiar with l'Hopital's Rule?

Or you could just use plain old integration by parts to get $\left[ x \ln (x) - x \right]_t^1 = -1 - t \ln (t) + t$.

Then calculate $\lim_{t \rightarrow 0^+} t \ln t = \lim_{t \rightarrow 0^+} \frac{\ln t}{\frac{1}{t}}$ (using l'Hopital's rule is the easiest way, which is why I asked if you were familiar with it):

$\lim_{t \rightarrow 0^+} \frac{\ln t}{\frac{1}{t}} = \lim_{t \rightarrow 0^+} \frac{\frac{1}{t}}{-\frac{1}{t^2}} = \lim_{t \rightarrow 0^+} (-t) = 0$.

So $-1 - \lim_{t \rightarrow 0^+}(t \ln (t) - t) = -1 - 0 = -1$

and therefore the integral is convergent.
• February 3rd 2008, 02:35 PM
TrevorP
Yeah the name l'Hopital seems familiar but I don't quite know what it is. I'll try a similar technique for the next few questions I have.
• February 3rd 2008, 03:07 PM
Jhevon
Quote:

Originally Posted by TrevorP
Yeah the name l'Hopital seems familiar but I don't quite know what it is. I'll try a similar technique for the next few questions I have.

you won't be able to use the technique if you don't know how it works.

basically, L'Hopital's rule says: if $\lim_{x \to a} \frac {f(x)}{g(x)} \to \frac 00 \mbox{ or } \frac {\infty}{\infty}$, then we can find the limit by taking the derivative of the numerator and denominator (repeatedly if necessary). that is, in either case above, $\lim_{x \to a} \frac {f(x)}{g(x)} = \lim_{x \to a} \frac {f'(x)}{g'(x)}$
• February 3rd 2008, 03:58 PM
TrevorP
Ok so here's my next one and my attempt on it:

$\int_0^1 \frac{dx}{\sqrt{x + x^2}} \ \ \ \ \begin{array}{c} \sqrt{x + x^2} \rightarrow 0 \ \ \mbox{as} \ \ x \rightarrow 0 \\ \sqrt{x + x^2} \rightarrow \sqrt{2} \ \ \mbox{as} \ \ x \rightarrow 1 \end{array}$

So there must be a singularity at x = 0.

Also:

$\frac{1}{\sqrt{x + x^2}} \sim \frac{1}{\sqrt{x}} \ \ \ \mbox{as x is small (approaches 0)}$

So:

$\int_0^1 \frac{dx}{\sqrt{x}} = \left. 2x^{\tfrac{1}{2}} \right|_0^1 < \infty \ \ \boxed{\mbox{Convergent}}$

Does this work???
• February 3rd 2008, 05:11 PM
ThePerfectHacker
Quote:

Originally Posted by TrevorP

$\int_0^1 \frac{dx}{\sqrt{x}} \ \ \ \ \begin{array}{c c} x = 1 + \tan^2 \theta \\ dx = 2 \tan \theta \sec^2 \theta d \theta \end{array}$

$2 \int_0^1 \tan \theta \sec \theta d \theta = \left. 2 \sec \theta \right|_0^1 < \infty$

$\int_0^1 \frac{dx}{\sqrt{x}} = \int_0^1 x^{-1/2} dx = 2 x^{1/2} \big|_{0^+}^1$
• February 3rd 2008, 06:13 PM
TrevorP
Ohh jeez. Thanks. So my logic makes sense then? Have I proved it's convergent?
• February 3rd 2008, 08:48 PM
mr fantastic
Quote:

Originally Posted by TrevorP
Ok so here's my next one and my attempt on it:

$\int_0^1 \frac{dx}{\sqrt{x + x^2}} \ \ \ \ \begin{array}{c} \sqrt{x + x^2} \rightarrow 0 \ \ \mbox{as} \ \ x \rightarrow 0 \\ \sqrt{x + x^2} \rightarrow \sqrt{2} \ \ \mbox{as} \ \ x \rightarrow 1 \end{array}$

So there must be a singularity at x = 0.

Also:

$\frac{1}{\sqrt{x + x^2}} \sim \frac{1}{\sqrt{x}} \ \ \ \mbox{as x is small (approaches 0)}$

So:

$\int_0^1 \frac{dx}{\sqrt{x}} = \left. 2x^{\tfrac{1}{2}} \right|_0^1 < \infty \ \ \boxed{\mbox{Convergent}}$

Does this work???

Quote:

Originally Posted by TrevorP
So my logic makes sense then? Have I proved it's convergent?

Yeah ..... probably ...... personally, I'd tighten the argument:

I'd make the observation that

$\frac{1}{\sqrt{x + x^2}} \leq \frac{1}{\sqrt{x}}$ for $0 < x \leq 1$

and so

$\int_0^1 \frac{1}{\sqrt{x + x^2}} \, dx \leq \int_0^1 \frac{1}{\sqrt{x}} \, dx$.

Therefore $\int_0^1 \frac{1}{\sqrt{x + x^2}} \, dx \leq \int_0^1 \frac{1}{\sqrt{x}} \, dx = 2[\sqrt{x}]_0^1 = 2$

and so is convergent.
• February 3rd 2008, 10:44 PM
mr fantastic
Quote:

Originally Posted by Jhevon
you won't be able to use the technique if you don't know how it works.

basically, L'Hopital's rule says: if $\lim_{x \to a} \frac {f(x)}{g(x)} \to \frac 00 \mbox{ or } \frac {\infty}{\infty}$, then we can find the limit by taking the derivative of the numerator and denominator (repeatedly if necessary). that is, in either case above, $\lim_{x \to a} \frac {f(x)}{g(x)} = \lim_{x \to a} \frac {f'(x)}{g'(x)}$

Thanks for that useful thumbnail, Jhevon. I'll just add the following:

$
\lim_{t \rightarrow 0^+} t \ln t$
has the indeterminant form $(0) \times (+\infty)$. The re-write to $\lim_{t \rightarrow 0^+} \frac{\ln t}{\frac{1}{t}}$ puts it into a 'standard' indeterminant $\frac {+\infty}{+\infty}$ form.

Note that ' $\, 0 \times \infty = \frac{1}{\infty} \times \infty = \frac{\infty}{\infty} \,$ '

(and you can count how many mathematical cows I've slaughtered here. But that's OK - slaughtering mathematical cows is something physicists do best ;) )