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Math Help - Convergent/Divergent Integrals

  1. #1
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    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}}
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    Quote Originally Posted by TrevorP View Post
    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?
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  3. #3
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    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}<br />
{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}<br />
{{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} }}<br />
{{k + 1}}} .

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

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

    The last sum is a telescoping series.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    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.
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  5. #5
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    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.
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    Quote Originally Posted by TrevorP View Post
    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)}
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    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???
    Last edited by TrevorP; February 3rd 2008 at 06:28 PM. Reason: edited for stupidity.
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    Quote Originally Posted by TrevorP View Post

    \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
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    Ohh jeez. Thanks. So my logic makes sense then? Have I proved it's convergent?
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  10. #10
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    Quote Originally Posted by TrevorP View Post
    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 View Post
    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.
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  11. #11
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    Quote Originally Posted by Jhevon View Post
    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:

    <br />
\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 )
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