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    Solution 169

    \displaystyle \int_0^{\frac{\pi}{2}} x\cot x\,dx=-\int_0^{\frac{\pi}{2}}\ln \sin x=\frac{\pi \ln 2}{2}

    Solution 170

    Let t=\sin^2  \theta:

    \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2x-1}\theta \cos^{2y-1}\theta \,d\theta =\frac{1}{2}\int_0^1 t^{x-1}(1-t)^{y-1}\,dt=\frac{1}{2}\,\text{B} \left(x,y\right)=\frac{\Gamma (x)\Gamma (y)}{2\,\Gamma (x+y)}

    Using the infinite product def. for \Gamma:

    \displaystyle\begin{aligned} \int_0^{\frac{\pi}{2}} \sin^{2x-1}\theta \cos^{2y-1}\theta \,d\theta &=\left(\frac{1}{2xy} \right) \left( \frac{1}{x+y} \right)^{-1} \frac{\displaystyle\prod_{n\geq 1}\frac{(1+n^{-1})^x}{1+xn^{-1}} \prod_{n\geq 1}\frac{(1+n^{-1})^y}{1+yn^{-1}}}{\displaystyle\prod_{n\geq 1}\left(\frac{(1+n^{-1})^{x+y}}{1+(x+y)n^{-1}}\right)}\\&= \frac{x+y}{2xy} \prod_{n\geq 1} \left(1+ \frac{x+y}{n} \right) \left(1+\frac{y}{n} \right)^{-1} \left(1+ \frac{x}{n} \right)^{-1}\\&=\frac{x+y}{2xy}\prod_{n \geq 1} \left(1+\frac{xy}{n(x+y+n)} \right)^{-1}\end{aligned}



    I'll leave the trig problem for those who wanted a (*) problem.

    (Original post by Jkn)
    But \displaystyle f'(\lambda)=\int_0^{\infty} \lambda \frac{x^{\lambda-1}\,dx}{(x^3+1)^2  }
    Bro do you lift? f'(\lambda) =x^{\lambda} \ln x
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    (Original post by Lord of the Flies)
    Solution 169

    \displaystyle \int_0^{\frac{\pi}{2}} x\cot x\,dx=-\int_0^{\frac{\pi}{2}}\ln \sin x=\frac{\pi \ln 2}{2}
    I love how you've shown all your working

    Seriously, if I ever see you, don't take the first 15 minutes of swearing personally.
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    (Original post by Zakee)
    Problem 169*/** (Finally a question that normal people can do without Stokes' theorem or Lagrange multipliers).

    This is a short but simple problem for you guys (your mathematical ability is far more superior to mine). Did take me a fair time though to realize what to do, but it was nice.

    Prove the following result:

    





\cos \frac{\pi}{7} - \cos \frac{2\pi}{7}+\cos \frac{3\pi}{7} = \frac{1}{2}.
    Solution 171:

     cos(\pi - x) = cos \pi cosx + sin \pi sinx

     cos(\pi - x) = -cosx

    So we have:

     cos\frac{\pi }{7} = - cos\frac{6\pi }{7}

     cos\frac{2\pi }{7} = -cos\frac{5\pi }{7}

     cos\frac{3\pi }{7} = -cos\frac{4\pi }{7}

    So we have:

     \cos \frac{\pi}{7} - \cos \frac{2\pi}{7}+\cos \frac{3\pi}{7} = \cos \frac{\pi}{7} + \cos \frac{3\pi}{7}+\cos \frac{5\pi}{7}

    We can then evaluate this using De Moivre's theorem:

     \cos \frac{\pi}{7} + \cos \frac{3\pi}{7}+\cos \frac{5\pi}{7} = \Re ( e^{\frac{i \pi }{7}}(1+e^\frac{i 2 \pi }{7}+e^\frac{i 4\pi }{7} ))

     = \Re ( e^\frac{i \pi }{7}( \dfrac{1-e^\frac{i 6\pi }{7}}{1-e^\frac{i 2\pi }{7}}))

     = \Re ( \dfrac{-1+e^\frac{i 6\pi }{7}}{e^\frac{i \pi }{7}-e^\frac{i -\pi }{7}}))

     = \Re ( \dfrac{-1-cos\frac{i \pi }{7}+isin\frac{i \pi }{7}}{2isin\frac{\pi }{7}} )

    =  \frac{1}{2}
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    (Original post by bananarama2)
    I love how you've shown all your working

    Seriously, if I ever see you, don't take the first 15 minutes of swearing personally.
    The first part is IBP using u = x and dv/dx = cotx. The second part is more complicated (A couple of  \frac{\pi }{2} - x substitutions in order to obtain the result, has come up in STEP before) but I'll type it up if you want it.
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    (Original post by DJMayes)
    The first part is IBP using u = x and dv/dx = cotx. The second part is more complicated (A couple of  \frac{\pi }{2} - x substitutions in order to obtain the result, has come up in STEP before) but I'll type it up if you want it.
    Pretty similar to solution 35...
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    (Original post by DJMayes)
    The first part is IBP using u = x and dv/dx = cotx. The second part is more complicated (A couple of  \frac{\pi }{2} - x substitutions in order to obtain the result, has come up in STEP before) but I'll type it up if you want it.
    I see how to do it, it's just it's not overly clear

    Edit: but thankyou
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    (Original post by shamika)
    LOLOLOLOLOLOL!

    I actually don't understand the sentence though. If it's referring to the Gaussian error integral, surely the authors do know how Mathematica (successfully) computes the integral? :confused::confused::confused:
    I think it (and the comment about it being in a collection of common integrals) is referring to the integral posed as the question.
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    (Original post by bananarama2)
    Seriously, if I ever see you, don't take the first 15 minutes of swearing personally.
    LOL.

    Something like this?

    Spoiler:
    Show
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    (Original post by Lord of the Flies)
    Solution 169

    \displaystyle \int_0^{\frac{\pi}{2}} x\cot x\,dx=-\int_0^{\frac{\pi}{2}}\ln \sin x=\frac{\pi \ln 2}{2}
    But the difficulty of the question lies in the fact that, if you use integration by parts, you have \left[x \ln(\sin x) \right]_0^{\frac{\pi}{2}} which requires some thought to evaluate.

    You must note that \displaystyle \lim_{x \to 0} x \ln(\sin x) = \lim_{x \to 0} x \ln(x) = 0 which must be shown using calculus or some sort of clever approach.

    Alternatively, there is another really nice approach that avoids this altogether!

    Whilst it is, given its close links to other questions recently posted, acceptable to quote \int_0^{\frac{\pi}{2}} \ln(\sin x) dx as a standard result, you need to remember that the things that you are omitting are the same sort of things that you get on STEP questions (i.e. using standard techniques and then reaching a roadblock) :lol:

    That said, I have enjoyed using many of your solutions to learn new techniques! Especially since much of the work is "left for the reader" :lol:
    Bro do you lift? f'(\lambda) =x^{\lambda} \ln x
    :facepalm: ****s sake, I should really hit the weights more...
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    (Original post by Jkn)
    You must note that \displaystyle \lim_{x \to 0} x \ln(\sin x) = \lim_{x \to 0} x \ln(x) = 0 which must be shown using calculus or some sort of clever approach.
    I would be comfortable quoting that sort of thing anywhere, except in a first year analysis exam where the point is to make you prove everything rigorously. This kind of limit, where you are essentially using the fact that a logarithm is slower than a polynomial is common knowledge.

    By making a leap from one correct statement to another, you're not lacking rigour (although you can argue you're lacking clarity, which is different).
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    (Original post by DJMayes)
    Solution 171:

     cos(\pi - x) = cos \pi cosx + sin \pi sinx

     cos(\pi - x) = -cosx

    So we have:

     cos\frac{\pi }{7} = - cos\frac{6\pi }{7}

     cos\frac{2\pi }{7} = -cos\frac{5\pi }{7}

     cos\frac{3\pi }{7} = -cos\frac{4\pi }{7}

    So we have:

     \cos \frac{\pi}{7} - \cos \frac{2\pi}{7}+\cos \frac{3\pi}{7} = \cos \frac{\pi}{7} + \cos \frac{3\pi}{7}+\cos \frac{5\pi}{7}

    We can then evaluate this using De Moivre's theorem:

     \cos \frac{\pi}{7} + \cos \frac{3\pi}{7}+\cos \frac{5\pi}{7} = \Re ( e^{\frac{i \pi }{7}}(1+e^\frac{i 2 \pi }{7}+e^\frac{i 4\pi }{7} ))

     = \Re ( e^\frac{i \pi }{7}( \dfrac{1-e^\frac{i 6\pi }{7}}{1-e^\frac{i 2\pi }{7}}))

     = \Re ( \dfrac{-1-e^\frac{i \pi }{7}}{e^\frac{i \pi }{7}-e^\frac{i -\pi }{7}}))

     = \Re ( \dfrac{-1-cos\frac{i \pi }{7}-isin\frac{i \pi }{7}}{2isin\frac{\pi }{7}} )

    =  \frac{1}{2}


    Very nice. I took a slightly different approach (I've not met De Moivre's yet - year 12), so here's my solution:


    Using the same rules as above:


    \cos \frac{5\pi}{7} = -\cos \frac{2\pi}{7}


    Therefore, the equation becomes:


    \cos \frac{\pi}{7}+\cos \frac{3\pi}{7}+\cos \frac{5\pi}{7} = \frac{1}{2}.

    We can now allow (a substitution), such that:

    Y =\cos \frac{\pi}{7}+\cos \frac{3\pi}{7}+\cos \frac{5\pi}{7}

    Using the following identity:

    \cos\alpha\cdot\sin\beta = \frac{1}{2}(sin(\alpha+\beta)-sin(\alpha-\beta)).


    If we multiply Y by \sin \frac{\pi}{7}

    Using our identity, we achieve:

    Y\sin\frac{\pi}{7} =  \frac{1}{2}({\sin\frac{2\pi}{7} + (\sin(\frac{4\pi}{7})-\sin\frac{2\pi}{7}) + (\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7})})

    Thus:

     \frac{1}{2}({\sin\frac{2\pi}{7} + (\sin(\frac{4\pi}{7})-\sin\frac{2\pi}{7}) + (\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7})}) =  \frac{1}{2}(\sin\frac{6\pi}{7})

    Using sin(\pi - x) = sinx

     \frac{1}{2}(\sin\frac{6\pi}{7}) =  \frac{1}{2}(\sin\frac{\pi}{7})

    As LHS = RHS (after simplification)

    Y\sin\frac{\pi}{7} =  \frac{1}{2}(\sin\frac{\pi}{7})

    Dividing both sides by \sin\frac{\pi}{7}

    Y = \frac{1}{2}

    As Y =\cos \frac{\pi}{7}+\cos \frac{3\pi}{7}+\cos \frac{5\pi}{7}

    and \cos \frac{\pi}{7}+\cos \frac{3\pi}{7}+\cos \frac{5\pi}{7} = \cos \frac{\pi}{7}-\cos \frac{2\pi}{7}+\cos \frac{3\pi}{7}

    \cos \frac{\pi}{7}-\cos \frac{2\pi}{7}+\cos \frac{3\pi}{7} = \frac{1}{2}


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    (Original post by Jkn)
    They're nothing difficult! The justification/adjusting is often excruciatingly tedious and difficult but with theorem like the Leibniz Integral Rule (aka. Feynman Integration aka. "Differentiating under the integral sign") you need only apply it without much thought (though you must assure the function is continuous in the relevant range - as usual) so it's definitely worth learning! Wikipedia is a good place to start, that's where I found a lot of these problems in fact :lol:

    In a nut shell: \frac{\partial}{\partial \alpha} f(\alpha, x) is done by assuming all other variables except the subject of your partial differentiation are constant.
    Spoiler:
    Show
    e.g: \displaystyle\frac{\partial^2}{ \partial y^2} xy^2=\frac{\partial}{\partial y} 2xy=2x
    Doesn't seem to bad thanks.
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    (Original post by Jkn)
    But \displaystyle f'(\lambda)=\int_0^{\infty} \lambda \frac{x^{\lambda-1}\,dx}{(x^3+1)^2  } ? :confused: But we havent proved that this would be equal to the thing with the log in? Have I missed something? :confused:


    On your marks...
    ...
    Get set...
    ...

    Problem 169*/**/*** (Looks friendly... at first)

    Evaluate \displaystyle\int_0^{\frac{\pi}{  2}}x\cot(x) dx

    Problem 170​***

    Prove that \displaystyle \int_0^{\frac{\pi}{2}} (\sin \theta)^{2x-1} (\cos \theta)^{2y-1} d\theta = \frac{x+y}{2xy} \prod_{n=1}^{\infty} \left(1+\frac{xy}{n(x+y+n)} \right)^{-1}
    Problem 169 is STEP III 1991 Q7. In the STEP on you are lead through it a bit though i suppose.
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    (Original post by Lord of the Flies)
    Disprove my claim by counterexample then
    What about problem 140? :lol: Unless the Basel problem's on the A-level spec, though I do take your point, some of these things could - hypothetically be done, but emphasis on hypothetical :teehee:
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    (Original post by bogstandardname)
    Problem 169 is STEP III 1991 Q7. In the STEP on you are lead through it a bit though i suppose.
    (Original post by shamika)
    I would be comfortable quoting that sort of thing anywhere, except in a first year analysis exam where the point is to make you prove everything rigorously. This kind of limit, where you are essentially using the fact that a logarithm is slower than a polynomial is common knowledge.

    By making a leap from one correct statement to another, you're not lacking rigour (although you can argue you're lacking clarity, which is different).
    Oh I didn't know it was a STEP question! But yeah, I've just had a look and it kind of illustrates my point that a one-line solution is not particularly appropriate.

    I was sure someone would do the easier way but I was hoping someone would relish the opportunity to practice a bit of Feynman Integration (i.e. the **/*** solution)

    Solution 169 (2)

    Let \displaystyle f(a)=\int_0^{\frac{\pi}{2}} \frac{\arctan (a \tan(x))}{ \tan(x)} dx

    Let \tan(x)=u,

    \displaystyle \Rightarrow \frac{\partial}{\partial a} f(a) = \displaystyle\int_0^{\frac{\pi}{  2}} \frac{1}{1+(a \tan(x))^2} dx 

=\displaystyle \int_0^{\infty} \frac{1}{(1+(au)^2)(1+u^2)} du = \displaystyle \frac{1}{a^2-1} \left[a \arctan(au) - arctan(u) \right]_0^{\infty}

    Noting that f(0)=0,

    \displaystyle \Rightarrow \int_0^{\frac{\pi}{2}} x \cot(x) dx = f(1)= \int_0^1 \frac{\pi}{2(a+1)} da = \left[\frac{\pi}{2} ln|a+1| \right]_0^1 = \frac{1}{2} \pi \ln(2) \ \square
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    (Original post by Jkn)
    But the difficulty of the question lies in the fact that, if you use integration by parts, you have \left[x \ln(\sin x) \right]_0^{\frac{\pi}{2}} which requires some thought to evaluate.
    That's where the difficult of the question lies? Please. Also, I'm failing to see how "it is acceptable to quote" that integral but not this limit.

    Alternatively, there is another really nice approach that avoids this altogether!
    I know what you are referring to. If you're going to be picky and require justification for the limit above, then differentiating under the integral requires a lot more justification. So the only thing you've avoided is the easier path.

    And in any case, this thread is not an exam, it's for fun. So unless something is terribly obscure/incorrect, it's hardly worth pointing out.

    Edit: voilà, you've just posted exactly what I was referring to.

    (Original post by joostan)
    What about problem 140? :lol: Unless the Basel problem's on the A-level spec, though I do take your point, some of these things could - hypothetically be done, but emphasis on hypothetical.
    Alright granted, that should have had an additional star. Also, I agree with you and shamika - in the future I will mention it if the (*) is an especially tough route to take (this is the case with the last integral I posted for instance).
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    (Original post by Lord of the Flies)
    That's where the difficult of the question lies? Please. Also, I'm failing to see how "it is acceptable to quote" that integral but not this limit.

    I know what you are referring to. If you're going to be picky and require justification for the limit above, then differentiating under the integral requires a lot more justification. So the only thing you've avoided is the easier path.

    And in any case, this thread is not an exam, it's for fun. So unless something is terribly obscure/incorrect, it's hardly worth pointing out.
    Okay, okay.. :rolleyes:

    It's just you have assumed that things that are of the level of STEP questions are trivial! Just look at 1991, STEP III, 7 where you are actually led to the answer step-by-step. So, what you have omitted as trivial is supposed to be even harder than a STEP question...

    I know it doesn't matter particularly and I'm not that bothered, but I got the impression from reading what a lot of people had said that people started to lose interest in this thread around the same time people started posting questions and/or solutions that are hard to relate to! And since even understanding your solution requires the reader to have abilities beyond the level of the average STEP integration question surely suggests that this kind of thing people were referring to :lol:
    Edit: voilà, you've just posted exactly what I was referring to.
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    Some more (*) problems for people who want them:

    Problem 172*

    Is \displaystyle\prod_{r=0}^{100} r! a perfect square? If not, could we remove one of the factorials to make it a perfect square?

    Problem 173*

    Let P(x) = 24x^{24} + \displaystyle\sum_{j=1}^{23} (24 - j) (x^{24-j} + x^{24+j} )

    Let z_{1}, z_{2}, ... , z_{r} be the distinct zeros of P(x) and let z_{k}^{2} = a_{k} + i b_{k}, \ k = 1, 2, ..., r and i = \sqrt{-1} and a, \ b \in \mathbb{R}

    Let \displaystyle\sum_{k=1}^{r} |b_{k} | = m + n \sqrt{p}

    where m, n, p are integers and p is not divisible by the square of any primes. Calculate m + n + p
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    (Original post by Felix Felicis)
    Some more (*) problems for people who want them:

    Problem 171*

    f(x) = 6x^{7} \sin^{2} \left( x^{1000} \right) e^{x^{2}}

    Find f^{2013} (0) (note f^{n} (x) is the nth derivative of f(x))
    Hasn't this already been posted by LOTF somewhere? :confused:
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    (Original post by ukdragon37)
    Hasn't this already been posted by LOTF somewhere? :confused:
    Damn! I didn't actually realise that, but yeah, it has
 
 
 
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