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    Any good method for this?
    (please attach I do not have home-works any more!!!)

    I have the correct answer (which is part of an undergrad question to which I am writing a solution)
    It is just that the method seems a bit too long and I do not want to rewrite the solution as this integral lies somewhere in the middle of a longish problem

    Thanks
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    TeeEm needing help?!? Thought I'd never see the day! :P
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    Integration by parts?
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    (Original post by iMacJack)
    TeeEm needing help?!? Thought I'd never see the day! :P
    happens to me more regularly than I would like to admit ...

    However in this question I have done the integral without problems; I just look for a quicker alternative if it exists.
    ( it is just that I am doing high level maths non-stop since this morning and my brain is now stalling )
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    (Original post by ubisoft)
    Integration by parts?
    I second that idea...and so does Wolfram Alpha 🙂
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    (Original post by TeeEm)
    happens to me more regularly than I would like to admit ...

    However in this question I have done the integral without problems; I just look for a quicker alternative if it exists.
    ( it is just that I am doing high level maths non-stop since this morning and my brain is now stalling )
    Lol! Fair enough man. My brain gets fried from too much FP1, so you're not doing too badly :P Keep up the good work man
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    (Original post by Gregorius)
    I second that idea...and so does Wolfram Alpha 🙂
    I split the log

    the second part I left alone

    The first part of the log
    I did substitution x = 2tantheta for the first part of the log
    then parts

    then combined all into the answer


    EDIT No need for substitution/ direct parts cuts the work a bit/ good enough
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    (Original post by TeeEm)
    I split the log
    Mathematical lumberjack as well now!
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    (Original post by Gregorius)
    Mathematical lumberjack as well now!
    I cut a bit of the work out now, so it is ok...

    It is actually one of the questions you posed the other night, that I took and made a question out of ...

    Find

    INTEGRAL (sin2x/x2) from 0 to infinity, using Laplace transform techniques
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    (Original post by Gregorius)
    Mathematical lumberjack as well now!
    (Original post by TeeEm)
    I cut a bit of the work out now, so it is ok...

    It is actually one of the questions you posed the other night, that I took and made a question out of ...

    Find

    INTEGRAL (sin2x/x2) from 0 to infinity, using Laplace transform techniques
    This reminds me, Greogrius - I tried computing a recurrence for the integral of nth powers of the since function, i.e a closed form for \displaystyle \int_0^{\infty} \left(\frac{\sin x}{x}\right)^n \, \mathrm{d}x and eventually gave up (although I managed to get the first few nth powers, up to 5 or so) - then a quick math.stackexchange lookup yielded and impressive and complex-looking closed form!
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    (Original post by Zacken)
    This reminds me, Greogrius - I tried computing a recurrence for the integral of nth powers of the since function, i.e a closed form for \displaystyle \int_0^{\infty} \left(\frac{\sin x}{x}\right)^n \, \mathrm{d}x and eventually gave up (although I managed to get the first few nth powers, up to 5 or so) - then a quick math.stackexchange lookup yielded and impressive and complex-looking closed form!
    I just borrowed the n=2 case to make a question for my LaplaceTransform resources
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    (Original post by TeeEm)
    I just borrowed the n=2 case to make a question for my LaplaceTransform resources
    I saw - I really need to brush up on my Laplace Transforms, perhaps starting with your resources! :-)
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    (Original post by Zacken)
    This reminds me, Greogrius - I tried computing a recurrence for the integral of nth powers of the since function, i.e a closed form for \displaystyle \int_0^{\infty} \left(\frac{\sin x}{x}\right)^n \, \mathrm{d}x and eventually gave up (although I managed to get the first few nth powers, up to 5 or so) - then a quick math.stackexchange lookup yielded and impressive and complex-looking closed form!
    Ooh! Do you have a link to the math.stackexchange article?
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    (Original post by Gregorius)
    Ooh! Do you have a link to the math.stackexchange article?
    Here you go, although most of the information/maths is in the PDFs/papers that they've linked to in the comments. :-)
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    (Original post by TeeEm)
    Any good method for this?
    (please attach I do not have home-works any more!!!)

    I have the correct answer (which is part of an undergrad question to which I am writing a solution)
    It is just that the method seems a bit too long and I do not want to rewrite the solution as this integral lies somewhere in the middle of a longish problem

    Thanks
    I=\int \log(\frac{x^2+1}{x^2}) \ dx

    Let x=\tan u \Rightarrow dx=\sec^2 u du

    So:

    I = \int \log(\frac{\sec^2 u}{\tan^2 u}) \ du = \int \log(\frac{1}{\sin^2 u}) \sec^2 u \ du = -2 \int \log(\sin u) \sec^2 u \ du

    Now by parts we have:

    I= -2 \{ \tan u \log(\sin u) - \int \tan u \cot u \ du \} = -2 \{ \tan u \log(\sin u) - u \} + C

    Now x=\tan u \Rightarrow \sin u = \frac{x}{\sqrt{x^2+1}} so:

    I = -2 \{ x \log(\frac{x}{\sqrt{x^2+1}}) - \tan^{-1} x \} + C = x \log(\frac{x^2+1}{x^2}) + 2 \tan^{-1} x + C

    (Confirmed by Wolfram)
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    (Original post by Zacken)
    I saw - I really need to brush up on my Laplace Transforms, perhaps starting with your resources! :-)
    I wrote many questions last week as I am helping 2 students at present,

    wait until I update the site as none of them are there

    here is the one I made tonight which lead to this integral
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    (Original post by Zacken)
    Here you go, although most of the information/maths is in the PDFs/papers that they've linked to in the comments. :-)
    Excellent! An endless stream of exam questions for the future!
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    (Original post by Gregorius)
    Excellent! An endless stream of exam questions for the future!
    Poor students.
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    (Original post by TeeEm)
    I wrote many questions last week as I am helping 2 students at present,

    wait until I update the site as none of them are there

    here is the one I made tonight which lead to this integral
    Thanks for this! :-)
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    (Original post by atsruser)
    I=\int \log(\frac{x^2+1}{x^2}) \ dx

    Let x=\tan u \Rightarrow dx=\sec^2 u du

    So:

    I = \int \log(\frac{\sec^2 u}{\tan^2 u}) \ du = \int \log(\frac{1}{\sin^2 u}) \sec^2 u \ du = -2 \int \log(\sin u) \sec^2 u \ du

    Now by parts we have:

    I= -2 \{ \tan u \log(\sin u) - \int \tan u \cot u \ du \} = -2 \{ \tan u \log(\sin u) - u \} + C

    Now x=\tan u \Rightarrow \sin u = \frac{x}{\sqrt{x^2+1}} so:

    I = -2 \{ x \log(\frac{x}{\sqrt{x^2+1}}) - \tan^{-1} x \} + C = x \log(\frac{x^2+1}{x^2}) + 2 \tan^{-1} x + C

    (Confirmed by Wolfram)
    Thank you

    I originally did that way but I did it directly by parts ( a tiny bit quicker)
    I was just wondering if there was an obvious method (as I am very tired) and this was part of a solution so I did not want to redo it at a later stage.
 
 
 
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