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    (Original post by Jkn)
    Problem 269***

    Using the previous problem as a hint,

    Find functions f and g such that \displaystyle \zeta(s)=f(s)+\int_0^{\infty} g(s,t) \ dt
    Dare I give:
    g(s,t) = 0, f(s) = \zeta(s)?
    Apologies if I've misunderstood \zeta - I flee from contour integration at every possible opportunity (not that there have been many such opportunities - I really mean that any time it might have been useful to learn contour integration, I ran away from it like the wind).
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    (Original post by Smaug123)
    Is this possible without contour integration? Contour integration is horrible
    Hmm, I'd imagine it would be doable with Taylor's series? Not too sure.
    (Original post by Smaug123)
    Dare I give:
    g(s,t) = 0, f(s) = \zeta(s)?
    Apologies if I've misunderstood \zeta - I flee from contour integration at every possible opportunity (not that there have been many such opportunities - I really mean that any time it might have been useful to learn contour integration, I ran away from it like the wind).
    I've modified the question now - the point of the question being to express the mysterious Zeta function in terms of less-mysterious 'elementary' functions which can, in turn, be used to approximate \zeta(2n+1) (n \in \mathbb{N}) by using integral approximations

    I learnt it this morning but I'm still feeling a bit shaky (hoping people will engage in discussion on ASOM with me with regards to a few problems!) I'm probably not quite at the point where I could have done the problem LOTF just solved, though I did think it worth posting as I knew a solution would not take long to emerge.
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    Problem 270

    what is: \displaystyle\int \frac{1}{x^{4}(1+x^{2})^{1/2}}dx ?
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    (Original post by Hasufel)
    Problem 270

    what is: \displaystyle\int \frac{1}{x^{4}(1+x^{2})^{1/2}}dx ?
    Solution 270

    \displaystyle \int \frac{dx}{x^{4} (1+x^2)^{\frac{1}{2}}} \overset{x = \tan t}= \int \cot^{3} t \csc t \ dt

    Now, note that \cot^{2} t + 1 \equiv \csc^{2} t

    \displaystyle \begin{aligned} \int \cot^{3} t \csc t \ dt = \int \cot t \csc t ( \csc^2 t - 1) \ dt &\overset{u = \csc t}= \int 1 - u^{2} \ du \\ & = u - \frac{1}{3} u^{3} + \mathcal{C} \\ & = \csc t - \frac{1}{3} \csc^{3} t + \mathcal{C} \\ & = \frac{\sqrt{x^{2} + 1}}{x} - \frac{1}{3} \frac{(x^2 + 1)^{\frac{3}{2}}}{x^3} + \mathcal{C} \\ & = \frac{\sqrt{x^2 + 1} (2x^2 - 1)}{3x^2} + \mathcal{C} \end{aligned}
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    (Original post by Felix Felicis)
    x
    Basically the same as yours but more direct:

    Solution 270 (2)

    Let x=\frac{1}{\sqrt{u^2-1}},

    \displaystyle \Rightarrow \int \frac{1}{x^4 \sqrt{1+x^2}} \ dx = \int (u^2-1)^2 \frac{\sqrt{u^2-1}}{u} \frac{-u}{(u^2-1)^{\frac{3}{2}}} \ du = \int 1- u^2 \ du 

\displaystyle = u - \frac{1}{3} u^3 + C=\frac{\sqrt{x^2+1} (2x^2-1)}{3x^2} + C

    Spoiler:
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    Whilst the substitution may seem arbitrary, it is best spotted by seekng a substitution that has that property that it combines powers of x, the derivative and \sqrt{1+x^2}. The first two will combine nicely regardless but the third is more difficult to find (we must seek the inverse). Our substation works nicely as f(x)=\frac{1}{\sqrt{x^2-1}} \Rightarrow \sqrt{1+(f(x))^2}=\frac{x}{\sqrt  {x^2-1}}.
    We can generalise by using the same substitution,

    \displaystyle \Rightarrow \int \frac{1}{x^{2(n+1)} \sqrt{1+x^2}} \ dx = - \int (u^2-1)^n \ du = - \sum_{r=0}^n \binom{n}{r} (-1)^{n-r} \int u^{2r} \ du 

\displaystyle \begin{aligned} = - \sum_{r=0}^n \binom{n}{r} (-1)^{n-r} \frac{u^{2r+1}}{2r+1} + C = - \sum_{r=0}^n \binom{n}{r} (-1)^{n-r} \frac{(1+x^2)^{ \frac{2r+1}{2} }}{x^{2r+1}(2r+1)} + C \end{aligned}
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    \displaystyle \gamma := \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k}-\log(n) \right) which is known as the Euler-Mascheroni constant.

    Problem 271
    ***

    Prove that \displaystyle \int_0^{\infty} e^{-x} \log(x) \ dx = \Gamma'(1) =\psi(1) = -\gamma.

    Note that I require you to prove any non-* theorems/results that you use in your proof.

    Problem 272*

    Prove that \displaystyle \begin{aligned} \gamma = -4 \int_0^{\infty} e^{-x^2} x \log(x) \ dx = -\int_0^1 \log \log \left(\frac{1}{x} \right) \ dx = \int_0^{\infty} \left( \frac{1}{e^x-1}-\frac{1}{xe^x} \right) \ dx \end{aligned}

\displaystyle = \int_0^1 \left( \frac{1}{\log(x)}+\frac{1}{1-x} \right) \ dx = \int_0^{\infty} \frac{1}{x} \left( \frac{1}{1+x^k} - e^{-x} \right) \ dx

    Problem 273***

    Evaluate \displaystyle \int_0^1 \int_0^1 \frac{x-1}{(1-xy) \log(xy)} \ dx

    Prove all non-* theorems, as before.

    Problem 274***

    Evaluate \displaystyle \int_0^{\infty} e^{-x^2} \log(x) \ dx

    Problem 275***

    Evaluate \displaystyle \int_0^{\infty} \frac{1}{x^2} \left( \frac{\log(1+x)}{(\log(x))^2+\pi  ^2} \right) \ dx
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    (Original post by Mladenov)
    Problem 165***

    Evaluate \displaystyle \int_{0}^{2} \frac{x^{4}}{(x^{2}+1)(x(2-x)^{3})^{\frac{1}{4}}}dx.
    Not sure this is right as my final answer looks messy, so let me know.

    Solution 165

    Let x \mapsto 2x:

    \displaystyle \int_0^2 \frac{(x^2+1)(x^2-1)+1}{(x^{2}+1)(x(2-x)^{3})^{\frac{1}{4}}} \ dx = \int_0^2 \left( \frac{x^2-1}{(x(2-x)^3)^{\frac{1}{4}}}+\frac{1}{(x  ^2+1)(x(2-x)^3)^{\frac{1}{4}}} \right) \ dx 

\displaystyle \underbrace{= \int_0^1 \frac{4x^2-1}{x^{\frac{1}{4}}(1-x)^{\frac{3}{4}}} \ dx}_{\alpha} + \underbrace{\int_0^1 \frac{1}{(4x^2+1)(x(1-x)^3)^{\frac{1}{4}}} \ dx}_{\beta}

    Comparing the integrals to the Beta function, using the Gamma function representation of the Beta function as well as Euler's reflection formula, we get:

    \displaystyle \alpha = 4 \int_0^1 x^{\frac{7}{4}}(1-x)^{-\frac{3}{4}} \ dx - \int_0^1 x^{-\frac{1}{4}} (1-x)^{-\frac{3}{4}} \ dx = 4 B \left(\frac{11}{4},\frac{1}{4} \right) - B \left(\frac{3}{4},\frac{1}{4} \right) 

\displaystyle = \frac{13}{8} \Gamma \left(1-\frac{1}{4} \right) \Gamma \left(\frac{1}{4} \right)=\frac{13 \pi}{8 \sin(\frac{\pi}{4})}=\frac{13 \pi \sqrt{2}}{8}

    Let x \mapsto \frac{1}{x+1} and then x \mapsto x^4:

    \displaystyle \beta = \int_0^{\infty} \frac{1}{(x+1)^2} \left( \frac{1}{(\frac{4}{(x+1)^2}+1)( \frac{x^3}{(x+1)^4})^{\frac{1}{4  }}} \right) \ dx = \int_0^{\infty} \frac{x+1}{x^{\frac{3}{4}}(x^2+2  x+5)} \ dx 

\displaystyle = 4 \int_0^{\infty} \frac{x^4+1}{x^8+2x^4+5} \ dx

    Splitting the denominator into its complex routes and integrating:

    \displaystyle \begin{aligned} = \frac{1}{2} \left[\sum_{ \{w : w^8+2w^4+5=0 \} } \frac{\log(x-w)}{w^3} \right]_0^{\infty} = -\frac{1}{2} \sum_{ \{w: w^8+2w^4+6=0 \} } \frac{\log(-w)}{w^3}=-\frac{\pi}{2} \left( \frac{1+i}{(-1-2i)^{\frac{3}{4}}} + \frac{1-i}{(-1+2i)^{\frac{3}{4}}} \right) \end{aligned}

    Simplifying the result as much as we can and combining \alpha and \beta, we get:

    \displaystyle \int_0^2 \frac{x^4}{(x^2+1)(x(2-x)^{3})^{\frac{1}{4}}} \ dx = \pi \left(\frac{13 \sqrt{2}}{8}+5^{-\frac{3}{4}} \left( \frac{2}{-11+15 \sqrt{5}-2 \sqrt{10(25-11 \sqrt{5})}} \right)^{-\frac{1}{4}} \right) \square
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    (Original post by Jkn)
    \displaystyle \gamma := \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k}-\log(n) \right) which is known as the Euler-Mascheroni constant.

    Problem 271
    ***

    Prove that \displaystyle \int_0^{\infty} e^{-x} \log(x) \ dx = \Gamma'(1) =\psi(1) = -\gamma.

    Note that I require you to prove any non-* theorems/results that you use in your proof.
    Does that include proof of differentiation under the integral sign?
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    (Original post by Felix Felicis)
    Does that include proof of differentiation under the integral sign?
    Sorry, no! I suppose I meant non-** theorems. Essentially I am asserting that \psi(1)=-\gamma is not quotable
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    Sort of a mixture between the last two solutions but I thought the route it leads down seems different enough to warrant a post:

    Solution 168
    (3)

    Beginning as in solution (1), we can deduce that
    \displaystyle \begin{aligned} \int_0^{\frac{\pi}{2}} \ln \sin x \ln \cos x \ dx = \frac{\pi}{2} \ln^2 2 - \frac{1}{6} \int_0^{\frac{\pi}{2}} \ln^2 \tan x \ dx \mathop =^{\tan x \mapsto x} \frac{\pi}{2} \ln^2 2- \frac{1}{6} \int_0^{\infty} \frac{\ln^2 x}{1+x^2} \ dx \end{aligned}

    Let \displaystyle \begin{aligned} I(\alpha) = \int_0^{\infty} \frac{x^{\alpha}}{1+x^2} \ dx \mathop =^{x^2 \mapsto x} \frac{1}{2} \int_0^{\infty} \frac{x^{\frac{\alpha-1}{2}}}{1+x} \ dx = \frac{1}{2} B \left(\frac{\alpha+1}{2}, \frac{1-\alpha}{2} \right) = \frac{1}{2} \Gamma \left( \frac{\alpha+1}{2} \right) \Gamma \left( 1-\frac{\alpha+1}{2} \right) \end{aligned}
    \displaystyle= \frac{\pi}{2} \csc \frac{\pi}{2} (\alpha+1)

    \displaystyle \begin{aligned} \Rightarrow \int_0^{\infty} \frac{x^{\alpha} \ln^2 x}{1+x^2} \ dx = \frac{\pi}{2} \frac{d^2}{d \alpha^2} \csc \frac{\pi}{2} (\alpha+1) = \frac{\pi^3}{8} \left(\csc \frac{\pi}{2} (\alpha+1) \cot^2 \frac{\pi}{2} (\alpha+1)+\csc^3 \frac{\pi}{2} (\alpha+1) \right) \end{aligned}

    \displaystyle \Rightarrow \int_0^{\frac{\pi}{2}} \ln \sin x \ln \cos x \ dx = \frac{\pi}{2} \ln^2 2 - \frac{1}{6} I''(0) = \frac{\pi}{2} \ln^2 2 - \frac{\pi^3}{48} \ \square

    We can generalise the last integral in a rather interesting way:

    \displaystyle \begin{aligned} \int_0^{\frac{\pi}{2}} \ln^n \tan x \ dx = \int_0^{\infty} \frac{\ln^n x}{1+x^2} \ dx = \frac{\pi}{2} \frac{d^n}{d \alpha^n} \csc \frac{\pi}{2} (\alpha+1) \Bigg|_{\alpha=0} = \frac{\pi^{n+1}}{2^{n+1}} \frac{d^n}{d \alpha^n} \csc \alpha \Bigg|_{\alpha=\frac{\pi}{2}} \end{aligned}

    We now note the Taylor's series of \csc x about x=\frac{\pi}{2}
    \displaystyle \csc x = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}E_{2k} \left(z-\frac{\pi}{2} \right)^{2k} for |z-\frac{\pi}{2}|<\frac{\pi}{2} where E_n denotes the nth Euler number.

    By considering the coefficients of this series, we can find closed forms for the above integral:

    \displaystyle \int_0^{\infty} \frac{\ln^n x}{1+x^2} \ dx = \frac{(-1)^{\frac{n}{2}} E_{n} \pi^{n+1}}{2^{n+1}} for all  n \in \mathbb{N}^{*}.

    Note that this implies that the above integral is equal to 0 whenever n is odd (this is implied both by the Taylor's series and the fact that all nth Euler numbers are 0 when n is odd).

    So, for example, \displaystyle \int_0^{\infty} \frac{\ln^8 x}{1+x^2} \ dx = \frac{1385 \pi^9}{2^9}

    Corollary: Letting n=0 gives us yet another way to deduce that \displaystyle \int_0^{\infty} \frac{1}{1+x^2} \ dx = \frac{E_0 \pi}{2}=\frac{\pi}{2}
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    Solution 274

    Let I = \displaystyle \int_{0}^{\infty} e^{-x^2} \ln x \ dx \overset{x \mapsto \sqrt{x}}= \frac{1}{4} \int_{0}^{\infty} x^{-\frac{1}{2}} e^{-x} \ln x \ dx

    Now, consider the gamma function:

    \displaystyle \begin{aligned} \Gamma (\lambda) = \int_{0}^{\infty} x^{\lambda - 1} e^{-x} \ dx  \implies \Gamma ' (\lambda ) & = \int_{0}^{\infty} \frac{\partial}{\partial \lambda} x^{\lambda - 1} e^{-x} \ dx \\ & = \int_{0}^{\infty} x^{\lambda - 1} e^{-x} \ln x \ dx \end{aligned}

    which evaluated at \lambda = \frac{1}{2} gives 4I

    We have that

    \displaystyle \begin{aligned} \psi (\lambda )  = \dfrac{\Gamma ' (\lambda)}{\Gamma ( \lambda )}  \implies \Gamma ' (\lambda ) & = \psi (\lambda ) \Gamma (\lambda ) \\  \implies \Gamma ' \left( \dfrac{1}{2} \right) & = \Gamma \left( \dfrac{1}{2} \right) \psi \left( \frac{1}{2} \right) \\ & = - \sqrt{\pi} \left( \gamma + 2 \ln 2 \right) = 4 I \end{aligned}

    \displaystyle \therefore \int_{0}^{\infty} e^{-x^2} \ln x \ dx = - \frac{1}{4} \sqrt{\pi} \left( \gamma + 2 \ln 2 \right)
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    (Original post by Felix Felicis)
    x
    Nice stuff man, don't you just love gamma functions, DUTIS and all that crap?

    Btw, do you like my two monster posts?
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    (Original post by Jkn)
    Nice stuff man, don't you just love gamma functions, DUTIS and all that crap?

    Btw, do you like my two monster posts?
    Gotta love it man, it's some good stuff :sogood:

    I do! Very nice I especially liked the extension to 168 And while I'm at it, I was looking through some old stuff and saw your method on analysing the Gaussian integral, that was quite nice too
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    (Original post by Felix Felicis)
    Gotta love it man, it's some good stuff :sogood:

    I do! Very nice I especially liked the extension to 168 And while I'm at it, I was looking through some old stuff and saw your method on analysing the Gaussian integral, that was quite nice too
    Ikr! You're going to absolutely destroy at uni next year. Are you applying (or should I saying going) to Cambridge and, if so, which college?

    Thanks man! I've got into the habit of extending everything now, it's so satisfying! I've fallen in love with Euler's reflection formula btw! I was messing about with the problem as I was trying to work through Mladenov's contour integration method, and then I though I would ass that to the mix. I then thought that I probably looked like a bit of a **** as I was bringing nothing new to the table and then realised that, whist repeated differentiation tends to be tedious, the Taylor's series was an awesome shortcut! We don't use Taylor's (or Laurent) series enough on this thread, perhaps you could dig up some fun questions?

    Cheers but I felt like a bit of an idiot right after I finished as I realised that it was entirely unnecessary given the fact that all I had to do was use a trigonometric substitution to trivially evaluate the appropriate value of the Beta function (though I left it there as the digamma method of evaluating that integral was interesting in itself ). Incidentally, it was that precise integral that was the entirety of STEP II Q2 2013 (which makes it really satisfying that I found it of my own accord)

    I'm looking forward to being taught how to evaluate the Gaussian next year :lol: I'll stroll up like "yeah I found like 10 ways to do this ever the summer :pierre:"

    I hope we are not being too narrow by being obsessed with integration and special functions..

    Oh, btw, if you liked some of those posts you might like the one I wrote about the E-M constant (etc..) a few pages back on ASOM
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    Problem 276

    (simple one, really)

    if u+v=50 which choice of both u and v make u \times v as large as posilble?
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    (Original post by Hasufel)
    Problem 276

    (simple one, really)

    if u+v=50 which choice of both u and v make u \times v as large as posilble?
    Two numbers which add to make 50 can be written in the form (25+k),(25-k) so that their product is:

     (25+k)(25-k) = 625 - k^2

    Clearly for this to be maximised k=0 (since a square is either positive or 0). Hence the largest product is 625.
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    (Original post by Hasufel)
    Problem 276

    (simple one, really)

    if u+v=50 which choice of both u and v make u \times v as large as posilble?
    Solution 276:
    I say:
    u=(25+ni) , v=(25-ni) where n is extraordinarily large.
    As:
    \lim_{n \to \infty}(25+ni)(25-ni) = \infty
    yet:
    u+v=50
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    (Original post by Hasufel)
    Problem 276

    (simple one, really)

    if u+v=50 which choice of both u and v make u \times v as large as posilble?
    Am I being silly? Can't you just change u \times v into u \times (50-u), then just maximize this to get 625?
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    (Original post by joostan)
    I say:
    u=(25+ni) , v=(25-ni) where n is extraordinarily large.
    As:
    \lim_{n \to \infty}(25+ni)(25-ni) = \infty
    yet:
    u+v=50
    I prefer to keep it real .
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    Problem 276 *

    Evaluate \displaystyle \int_0^{2\pi}\frac{1}{e^{\sin x}+1}\,dx

    Spoiler:
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    I found this in a Hong Kong A-level paper but they walked you through to a solution. If you want a big hint check out the spoiler.
    Spoiler:
    Show
    Consider a substitution of the form x \rightarrow \alpha-x for some suitable \alpha
 
 
 
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