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    (Original post by james22)
    How can you evaluate that integral though?
    Doesn't seem impossible, right? It's e^z around the unit semicircle. Currently practically asleep, though, so I might be talking nonsense.
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    (Original post by james22)
    How can you evaluate that integral though?
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    DUTIS
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     \displaystyle f(\alpha) = \int^{\pi}_{0} e^{\alpha e^{ix}} \ dx
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    (Original post by Smaug123)
    Doesn't seem impossible, right? It's e^z around the unit semicircle. Currently practically asleep, though, so I might be talking nonsense.
    Surely then the integral would be 0 (no poles), which is not correct?

    (I haven't done contour integrals in over a year btw.)
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    (Original post by james22)
    Surely then the integral would be 0 (no poles), which is not correct?

    (I haven't done contour integrals in over a year btw.)
    It's e^z around the unit semicircle, which isn't a closed loop. We just need to evaluate the integral along the x-axis to complete the semicircular closed contour: -\int_{-1}^1 e^z dz. OK, I've definitely got something horribly wrong :P
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    (Original post by Noble.)
    I can't be bothered to dig out some of my notes, so this may not be rigorous:
    Nice, I also got it into exponential form but didn't think writing it as a sum would lead anywhere
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    (Original post by Smaug123)
    It's e^z around the unit semicircle, which isn't a closed loop. We just need to evaluate the integral along the x-axis to complete the semicircular closed contour: -\int_{-1}^1 e^z dz. OK, I've definitely got something horribly wrong :P
    The answer is not real, and that integral is real, so something has gone very wrong. I cannot see waht though which is really worrying because I should be pretty good at complex analysis.
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    Solution 483

    This is an elementary solution to the problem (here meaning not using complex analysis). For full disclosure, I used ThatPerson's hint.

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     I = \displaystyle\int_{-\pi}^{\pi} \dfrac{e^{\sin(x)+\cos(x)}}{e^{\  sin(x)}+e^{x}} \cos(\sin(x))\ dx

    As in the previous problem (482) it is good to use symmetry; in this case the limits suggestion a substitution of the form  y = -x . Applying this and multiplying out denominators, we obtain:

     I =  \displaystyle\int_{-\pi}^{\pi} \dfrac{e^y+\cos(y)}{e^{\sin(y)}+  e^{y}} \cos(\sin(y))\ dy

    Re-labelling the variable and adding these two expressions together cancels out a lot of horrible stuff, yielding:

     2I = \displaystyle\int_{-\pi}^{\pi} e^{\cos(x)}\cos(\sin(x)) \ dx

    The integrand is even, and so we reduce to:

     I = \displaystyle\int_{0}^{\pi} e^{\cos(x)}\cos(\sin(x)) \ dx

    Using De Moivre's theorem twice, we can rewrite this as:

     I = \Re \displaystyle\int_{0}^{\pi} e^{e^{ix}} \ dx

    We solve this new integral as follows. Let:

     I[\alpha] = \displaystyle\int_{0}^{\pi} e^{\alpha e^{ix}} \ dx

     \Rightarrow \dfrac{dI}{d\alpha} = \displaystyle\int_{0}^{\pi} e^{ix}e^{\alpha e^{ix}} \ dx

     \Rightarrow \dfrac{dI}{d\alpha} = 2\frac{i\sinh(\alpha)}{\alpha}

    Solving this differential equation and applying the easy boundary condition  \alpha = 0 , we obtain:

     I[\alpha] = \pi + \Im

    Where  \Im ] is an imaginary function we are not interested in (though an explicit form can be found via DUTIS if we so desire).

    And so our final answer is  \pi .

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    (Original post by DJMayes)
    Solution 483

    This is an elementary solution to the problem (here meaning not using complex analysis). For full disclosure, I used ThatPerson's hint.

    Spoiler:
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     I = \displaystyle\int_{-\pi}^{\pi} \dfrac{e^{\sin(x)+\cos(x)}}{e^{\  sin(x)}+e^{x}} \cos(\sin(x))\ dx

    As in the previous problem (482) it is good to use symmetry; in this case the limits suggestion a substitution of the form  y = -x . Applying this and multiplying out denominators, we obtain:

     I =  \displaystyle\int_{-\pi}^{\pi} \dfrac{e^y+\cos(y)}{e^{\sin(y)}+  e^{y}} \cos(\sin(y))\ dy

    Re-labelling the variable and adding these two expressions together cancels out a lot of horrible stuff, yielding:

     2I = \displaystyle\int_{-\pi}^{\pi} e^{\cos(x)}\cos(\sin(x)) \ dx

    The integrand is even, and so we reduce to:

     I = \displaystyle\int_{0}^{\pi} e^{\cos(x)}\cos(\sin(x)) \ dx

    Using De Moivre's theorem twice, we can rewrite this as:

     I = \Re \displaystyle\int_{0}^{\pi} e^{e^{ix}} \ dx

    We solve this new integral as follows. Let:

     I[\alpha] = \displaystyle\int_{0}^{\pi} e^{\alpha e^{ix}} \ dx

     \Rightarrow \dfrac{dI}{d\alpha} = \displaystyle\int_{0}^{\pi} e^{ix}e^{\alpha e^{ix}} \ dx

     \Rightarrow \dfrac{dI}{d\alpha} = i\sinh(\alpha)

    Solving this differential equation and applying the easy boundary condition  \alpha = 0 , we obtain:

     I[\alpha] = \pi - i + i\cosh(\alpha)

    And so our final answer is  \pi .

    I'm not sure that
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      \displaystyle\int_{0}^{\pi} e^{ix}e^{\alpha e^{ix}} \ dx = i\sinh(\alpha)
    . I got a different answer for that part, though it doesn't affect the end since we're only interested in the real part.

    Also, apparently this was set in a Cambridge exam over a hundred years ago:

    Problem 484**

     \displaystyle \int^{4}_{0} \dfrac{\ln (x)}{\sqrt{4x-x^2}} \ dx
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    (Original post by ThatPerson)
    I'm not sure that
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      \displaystyle\int_{0}^{\pi} e^{ix}e^{\alpha e^{ix}} \ dx = i\sinh(\alpha)
    . I got a different answer for that part, though it doesn't affect the end since we're only interested in the real part.

    Also, apparently this was set in a Cambridge exam over a hundred years ago:

    Problem 484**

     \displaystyle \int^{4}_{0} \dfrac{\ln (x)}{\sqrt{4x-x^2}} \ dx
    Good call; I was being very careless with my constants. Will fix that.
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    (Original post by Flauta)
    Is it
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    4/e
    ?

    I'll type up my "solution" tomorrow if I'm right, I've only done a little probability though so the chances are I'm far off.
    Indeed it is
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    (Original post by ThatPerson)
    I'm not sure that
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      \displaystyle\int_{0}^{\pi} e^{ix}e^{\alpha e^{ix}} \ dx = i\sinh(\alpha)
    . I got a different answer for that part, though it doesn't affect the end since we're only interested in the real part.

    Also, apparently this was set in a Cambridge exam over a hundred years ago:

    Problem 484**

     \displaystyle \int^{4}_{0} \dfrac{\ln (x)}{\sqrt{4x-x^2}} \ dx
    I get my answer as:

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    0. This answer looks slightly dubious, however I cannot see any error with my method. :rolleyes: My 0 arises from \pi ln2 - \pi ln2
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    (Original post by newblood)
    I get my answer as:

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    0. This answer looks slightly dubious, however I cannot see any error with my method. :rolleyes: My 0 arises from \pi ln2 - \pi ln2
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    Your answer is correct.
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    (Original post by ThatPerson)
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    Your answer is correct.
    In which case, that's a neat little integral
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    (Original post by newblood)
    Problem 480 ** (not much probability theory in A-Level)

    Let  X_1, X_2,...,X_n be i.i.d r.v's with uniform distribution on  [1,2] .

    Determine a unique value,  \Delta , such that \lim_{n\to \infty} Pr(a &lt;(X_1X_2...X_n)^{1/n} &lt; b) = 1 iff [latex] a< \Delta
    Solution 480

    Name:  14202228825781009771754.jpg
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    There's another method involving how the arithmetic and geometric means are related
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    (Original post by jjpneed1)
    Pretty average tbh

    Problem 471
    Determine all sets of non-negative integers  x, y and z that satisfy  2^x + 3^y = z^2 .

    Problem 472
    Evaluate  \displaystyle \int_0^{\infty} \frac{\ln(1 + x)\ln(1 + x^2)}{x^3}  dx

    Last ones for a while now while I sit some tasty exams, gl on that integral, good job if someone gets it out before exams finish
    Just was looking back a few pages and saw 472. Does that integral have a nice answer? I plugged it into Mathematica and the indefinite integral is horrific.

    Edit: Surprisingly, I'm making some progress

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    Edit #2 The answer actually seems to be  \dfrac{\pi}{2} . I made a mistake earlier when simplifying.
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    Problem 485**

    Prove that every open set G on the real line is the union of a finite or countable system of pairwise disjoint open intervals, where we regard (-\infty,x),(-\infty,\infty),(y,\infty) as open intervals.
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    (Original post by ThatPerson)
    Just was looking back a few pages and saw 472. Does that integral have a nice answer? I plugged it into Mathematica and the indefinite integral is horrific.

    Edit: Surprisingly, I'm making some progress

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    I've hit a dead-end for now. Though I believe the answer is  \dfrac{5{\pi}^2}{48} , which makes me suspect series or the zeta function is involved as that expression is equal to  \dfrac{5}{8} \zeta{(2)} .
    Honesty I can't remember the answer, I think it's something like that though.
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    (Original post by 0x2a)
    Problem 485**

    Prove that every open set G on the real line is the union of a finite or countable system of pairwise disjoint open intervals, where we regard (-\infty,x),(-\infty,\infty),(y,\infty) as open intervals.
    Solution 485

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    Let  y \in G . Then by definition  \exists r&gt;0 such that  B(y, r) \subseteq G . Given this, we can easily construct a (possibly uncountable) set of open intervals, the union of which is G:

     \cup_{y \in G} B(y,r) = G

    Now, we take this set and construct a new one by the following method: If two intervals are not disjoint, then their union is an interval; take this union in all possible cases. The result is a set of pairwise disjoint open intervals equal to our set. It remains to prove that this set is countable; to do this we note that our set of intervals injects into the rationals and we are done.

    This questions feels very simple if you know the expected content but due to that I am wary I may have made a mistake here somewhere.

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    (Original post by 0x2a)
    Problem 485**

    Prove that every open set G on the real line is the union of a finite or countable system of pairwise disjoint open intervals, where we regard (-\infty,x),(-\infty,\infty),(y,\infty) as open intervals.
    For each x \in G, but a rational p_x in the same path connected component as x (existance is obvious from the definition of openness). For each p_x let a_x,b_x be such that (a_x,b_x) \subseteq G but a_x,b_x \not\in G (existance comes from G being open). We then have countable many open intervals, and they clearly cover all of G
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    (Original post by DJMayes)
    Solution 485

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    Let  y \in G . Then by definition  \exists r&gt;0 such that  B(y, r) \subseteq G . Given this, we can easily construct a (possibly uncountable) set of open intervals, the union of which is G:

     \cup_{y \in G} B(y,r) = G

    Now, we take this set and construct a new one by the following method: If two intervals are not disjoint, then their union is an interval; take this union in all possible cases. The result is a set of pairwise disjoint open intervals equal to our set. It remains to prove that this set is countable; to do this we note that our set of intervals injects into the rationals and we are done.

    This questions feels very simple if you know the expected content but due to that I am wary I may have made a mistake here somewhere.

    I'm not sure about this, since you seem to be trying to perform uncountably many operations. I think it needs more justification to be correct.
 
 
 
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