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Changing limits of integration to infinity Watch

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    I am doing the final part of the question, so using the substitution suggested:

    x = \tan \theta

    when \theta = \frac{\pi}{2}, x = \infty    (*)

    \Rightarrow when  \theta = \frac{3 \pi}{2}, x = \infty

    \Rightarrow \tan \frac{\pi}{2} = \tan \frac{3 \pi}{2}

    \Rightarrow \frac{1}{2} = \frac{3}{2}

    I don't understand how you can say this (*), surely it is undefined?
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    (Original post by lizard54142)
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    I am doing the final part of the question, so using the substitution suggested:

    x = \tan \theta

    when \theta = \frac{\pi}{2}, x = \infty    (*)

    \Rightarrow when  \theta = \frac{3 \pi}{2}, x = \infty

    \Rightarrow \tan \frac{\pi}{2} = \tan \frac{3 \pi}{2}

    \Rightarrow \frac{1}{2} = \frac{3}{2}

    I don't understand how you can say this (*), surely it is undefined?
    It's a very minor abuse of notation. If you restrict theta to the interval 0 to pi/2 then as theta approaches pi/2, x increases unboundedly i.e. it approaches positive infinity. We just write the infinity symbol as the upper limit even though it's not actually a number to represent what's really going on.
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    (Original post by davros)
    It's a very minor abuse of notation. If you restrict theta to the interval 0 to pi/2 then as theta approaches pi/2, x increases unboundedly i.e. it approaches positive infinity. We just write the infinity symbol as the upper limit even though it's not actually a number to represent what's really going on.
    Thanks for the reply.

    Okay, yeah that's what I originally thought. Say then we define an integral as:

    \displaystyle\int^a_0 f(x)\ dx

    let  u = \dfrac{1}{x}

    Using this substitution, how would the limits change in this case?
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    (Original post by lizard54142)
    Thanks for the reply.

    Okay, yeah that's what I originally thought. Say then we define an integral as:

    \displaystyle\int^a_0 f(x)\ dx

    let  u = \dfrac{1}{x}

    Using this substitution, how would the limits change in this case?
    You need to be careful because 1/x is again undefined at 0. If a > 0 then you're using that branch of 1/x which approaches +infinity as x->0 from above so again you will use +infinity as the limit of integration.
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    (Original post by davros)
    You need to be careful because 1/x is again undefined at 0. If a > 0 then you're using that branch of 1/x which approaches +infinity as x->0 from above so again you will use +infinity as the limit of integration.
    Okay, so the new integral would be:

    \displaystyle\int^\infty_\frac{1  }{a} \ f(u) \dfrac{x^2}{1} \ du

    Is this correct?

    Also another question, say we have evaluated a definite integral and have this:

    \displaystyle \left[ f(x) \right]_0^\infty

    Is this equal to:

    \displaystyle(\lim_{x\to \infty} f(x)) - f(0)
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    (Original post by lizard54142)
    Okay, so the new integral would be:

    \displaystyle\int^\infty_\frac{1  }{a} \ f(u) \dfrac{x^2}{1} \ du

    Is this correct?

    Also another question, say we have evaluated a definite integral and have this:

    \displaystyle \left[ f(x) \right]_0^\infty

    Is this equal to:

    \displaystyle(\lim_{x\to \infty} f(x)) - f(0)
    You need a 1/u^2 in your integral - you can't leave it as a mixture of u's and x's!

    And yes, your final expression is how we define how to evaluate a definite integral when one of the limits is infinite.
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    (Original post by davros)
    You need a 1/u^2 in your integral - you can't leave it as a mixture of u's and x's!

    And yes, your final expression is how we define how to evaluate a definite integral when one of the limits is infinite.
    Yeah I understand you can't have a mixture, I just left it like that because I was lazy okay I think I have this sorted now, thanks.
 
 
 
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