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Math Help - How Do I look up...

  1. #1
    MHF Contributor Mathstud28's Avatar
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    How Do I look up...

    or where do I learn how to do the introducing two integrals to eliminate the complexity of an integral? I mean how do you do it? Can someone spare a minute to fill me in?
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    I have to say that I really love that method, I personally use it in a variety of cases, but not always works.

    For example, compute \int_1^2\ln x\,dx. The straightforward sense tells us that we must apply integration by parts, but look another way. For x>0,\,\ln x=\int_1^x\frac1y\,dy, and there it is, we have our parameter to construct our double integral with the only reason to reverse integration order and make a little bit simple the problem.

    Another example, find \int_0^1\ln(1+\sqrt x)\,dx. Again, follow the same idea, construct a double integral and reverse integration order, you'll get a simple integral to tackle.

    How about the famous Dirichlet Integral \int_0^\infty\frac{\sin x}{x}\,dx? This can also be tackled by introducing \frac1x=\int_0^\infty e^{-xy}\,dy,\,x>0, the result follows... and so on.

    A lot of single integrals can be solved with double integration tricks, but it depends.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Ok

    So \int_0^1\ln(1+\sqrt x)\,dx= img.top {vertical-align:15%;} \int_0^x\frac{\1}{2\sqrt x(1+\sqrt x)}\,dx\,dx" alt="int_0^1\ \int_0^x\frac{\1}{2\sqrt x(1+\sqrt x)}\,dx\,dx" />? and if so where do you go from there?
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  4. #4
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    Krizalid's Avatar
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    Clarify please.
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    As you can see Krizalid likes to use double integrals, which really help a lot. I like to use infinite series. But those can be hard to sum at times. Complex analysis also helps, but again not always. You should learn different techniques for integration just in case on approach is really bad.

    Here is another example. I was talking with my classmates today about integration tricks and I mentioned double integration.

    Consider, \int_0^{\infty} \frac{\cos (bx) - \cos (ax)}{x^2} dx = \int_0^{\infty} \int_b^a \frac{\sin (yx)}{x} dy ~ dx.
    Now change order of integration and you will get, \frac{\pi}{2}(b-a) as the answer.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Ok

    But where do you go from there...now that you have the new integral how do you evaluate it...and why is it easier...a couple of examples I tried just reverted back to the original integral?
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    But where do you go from there...now that you have the new integral how do you evaluate it...and why is it easier...a couple of examples I tried just reverted back to the original integral?
    Well, have you ever learned double integration? Do you know how to take double integrals? That is the first step.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    I thought I did

    Dont you evaluate just evaluate them as normal one at a time and regard the x or y that isnt the integration variable(dx,dy) as a constant then integrate the other?


    And also how do you get them into these new forms I understood the \frac1x=\int_0^\infty e^{-xy}\,dy,\,x>0, but I dont understand how you converted your example...nor do I understand the Dirichlet application
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    Dont you evaluate just evaluate them as normal one at a time and regard the x or y that isnt the integration variable(dx,dy) as a constant then integrate the other?
    Exactly. But what the trick is to change the order of integration instead of integrating (dx,dy) you integrate (dy,dx).

    And also how do you get them into these new forms I understood the \frac1x=\int_0^\infty e^{-xy}\,dy,\,x>0, but I dont understand how you converted your example...nor do I understand the Dirichlet application
    Think over here of x as being just a fixed number. And use substitution.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Well

    \int_1^2\ln x\,dx\= \int_1^2\int_1^{x}\ln y\,dy\ so you get ]= \int_1^{x}\int_1^2\frac{1}{y}\,dy\= \int_1^{x}\ln(2)-ln(1)\= \int_1^{x}\ln(2))\= \ln(2)x-ln(2)\

    but I must have done something wrong because that doesnt make sense?
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  11. #11
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    Quote Originally Posted by Mathstud28 View Post
    \int_1^2\ln x\,dx\= \int_1^2\int_1^{x}\ln y\,dy\
    This makes no sense, besides, you can't reverse integration order as you did.

    See double integration topics first.
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Sorry

    I meant

    \int_0^1\ln x,dx\= \int_1^{x}\int_1^2\frac{1}{y}\,dx,dy\= \int_1^{x}\ln(2),dy= \ln(2) x-ln(2)\ so what did I do wrong?
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    Ok is this semi-right?

    \int_1^2\ln x\,dx\= \int_1^2\int_1^{x}\frac{1}{y}\,dy,dx\=]= \int_1^{x}\int_1^2\frac{1}{y}\,dx,dy\= \int_1^{x}\frac{2}{y}-\frac{1}{y}\,dy\=ln(x) what am I doing wrong will you point it out to me?
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \int_1^2\ln x\,dx\= \int_1^2\int_1^{x}\frac{1}{y}\,dy,dx\=]= \int_1^{x}\int_1^2\frac{1}{y}\,dx,dy\= \int_1^{x}\frac{2}{y}-\frac{1}{y}\,dy\=ln(x) what am I doing wrong will you point it out to me?
    graph the region you are integrating over so you know how to alter the limits of integration when swirching from dxdy to dydx

    \int_1^2 \ln x~dx = \int_1^2 \int_1^x \frac 1y~dydx = \int_1^2 \int_y^2 \frac 1y~dxdy
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  15. #15
    MHF Contributor Mathstud28's Avatar
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    Could

    you please explain what the indicators are on the graph....actually first off what do you graph and what indicators tell you what the limits of integration need to be changed to? thanks by the way
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