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    (Original post by Sketch)
    Is this the (***) solution? If not, I don't have any knowledge of double integrals at A Level! :lol:
    Pf! I'm trying to stay away from this thread because I need to work and you quote me back here! Yes it is. A similar method, again with double integrals, is to make all the limits independent, so that the switch is more intuitive:

    \displaystyle\begin{aligned}\int  _0^{\infty} \frac{\arctan ex-\arctan x}{x}\,dx &= \int_0^{\infty}\int_1^{e}\frac{  1}{1+t^2x^2}\,dt \,dx \\ &=\int_1^{e}\int_{0}^{\infty} \frac{1}{1+t^2x^2}\,dx \, dt=\int_1^{e}\frac{\pi}{2t}\,dt=  \frac{\pi}{2}\end{aligned}

    (Original post by Jkn)
    For ****s sake :lol: classic physicist :lol:
    Actually, it's a very standard switch for double integrals, and what he has written is perfectly rigorous. And I'm sorry for picking on you again, but it looks like you are commenting something you don't know much about - the substitution is not r^2=x^2+y^2 at all.
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    (Original post by Mladenov)
    As you want another proof for problem 128, try to prove the following result:

    Problem 130*

    \begin{aligned} \displaystyle \frac{n}{2n+1}\frac{((2n)!!)^{2}  }{((2n-1)!!)^{2}(2n+1)} < \left(\int_{0}^{\infty} e^{-x^{2}}dx \right)^{2} < \frac{n}{2n-1}\frac{(2n-3)!!)^{2}(2n-1)}{((2n-2)!!)^{2}}\frac{\pi}{2} \end{aligned}.
    Then note that e^{-x^{2}} is even.
    Well it's so obvious when you put it like that(!)
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    (Original post by Lord of the Flies)
    Actually, it's a very standard switch for double integrals, and what he has written is perfectly rigorous. And I'm sorry for picking on you again, but it looks like you are commenting something you don't know much about - the substitution is not r^2=x^2+y^2 at all.
    ****s sake hahaha :lol:

    Exactly, I have no idea what a double integral even is (asides from it meaning integrating something twice). But it would be nice for such a step to be justified in terms of regular integration Looking back on my post, I suppose what I'm really doing is requesting a proof/derivation of this "double integration" technique

    Oh. What is it then?
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    (Original post by Mladenov)
    As you want another proof for problem 128, try to prove the following result:

    Problem 130*

    \begin{aligned} \displaystyle \frac{n}{2n+1}\frac{((2n)!!)^{2}  }{((2n-1)!!)^{2}(2n+1)} < \left(\int_{0}^{\infty} e^{-x^{2}}dx \right)^{2} < \frac{n}{2n-1}\frac{(2n-3)!!)^{2}(2n-1)}{((2n-2)!!)^{2}}\frac{\pi}{2} \end{aligned}.
    Then note that e^{-x^{2}} is even.
    Hmm, looks suspiciously similar to the series I derived (it included double factorial notation). I'll give it a go :lol: (It doesn't require double integrals does it?)
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    (Original post by Jkn)
    Hmm, looks suspiciously similar to the series I derived (it included double factorial notation). I'll give it a go :lol: (It doesn't require double integrals does it?)
    It does not.
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    (Original post by Jkn)
    ****s sake hahaha :lol:

    Exactly, I have no idea what a double integral even is (asides from it meaning integrating something twice). But it would be nice for such a step to be justified in terms of regular integration Looking back on my post, I suppose what I'm really doing is requesting a proof/derivation of this "double integration" technique

    Oh. What is it then?
    Spoiler:
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    No worries

    Well, substitutions are not as simple with double integrals. When you substitute x=f(r,s) and y=g(r,s), you cannot simply take dx or dy as you usually would, because you are dealing with several variables. Roughly speaking, what happens it that you compute a set of partials, then find the determinant of a matrix whose elements are these partials, and voilà! For more information, google "Jacobian". There is a pretty good course on this in "MIT Open course ware", if you are interested (it is very accessible as well, and the lecturer is French!). In this case, the sub is:

    x=r\cos \theta

    y=r\sin \theta

    (can you now see why the substitution does indeed cover the entire real plane?)

    When doing the process described above, you will find that the Jacobian det is simply r , hence why there is an extra r in the integrand.
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    (Original post by Mladenov)
    It does not.
    Hmm.. seems a bit of a beast. I've simplified the LHS in terms of the maclaurin's arctan expansion so I can pretty much prove it. But were you suggesting I start with the middle? (Or trap it between the multiples of \pi? )
    (Original post by Lord of the Flies)
    Spoiler:
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    No worries

    Well, substitutions are not as simple with double integrals. When you substitute x=f(r,s) and y=g(r,s), you cannot simply take dx or dy as you usually would, because we are dealing with several variables. Roughly speaking, what happens it that you compute a set of partials, then find the determinant of a matrix whose elements are these partials, and voilà! For more information, google "Jacobian". There is a pretty good course on this in "MIT Open course ware", if you are interested (it is very accessible as well, and the lecturer is French!). In this case, the sub is:

    x=r\cos \theta

    y=r\sin \theta

    (can you now see why the substitution does indeed cover the entire real plane?)

    When doing the process described above, you will find that the Jacobian det is simply r , hence why there is an extra r in the integrand.
    Hmm, sounds interesting. I will add it to my post-STEP reading list

    Well yes it has always been obvious that they cover it but the equivalence isn't! Anyway... I need to stop asking people to rigorously prove things they regard as assumed knowledge (I have a habit of doing that :lol:)
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    (Original post by Jkn)
    ...
    Shall, I recommend you commence with e^{-x^{2}}.
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    I am not sure which was more of an adventure - doing the problem, or typing the solution.

    Solution 130

    First we prove an intermediate result:

    \displaystyle \begin{aligned}y_n=\int_0^{\pi/2}\cos^nt\,dt\Rightarrow y_n&=(n-1)\int_0^{\pi/2}\sin^2 t\cos^{n-2}t\,dt\\[7pt]&=(n-1)y_{n-2}-(n-1)y_{n}\\[7pt]&\qquad\Rightarrow y_n=\frac{n-1}{n}y_{n-2}\\[7pt]&\qquad\Rightarrow\left\{ \begin{array}{l l} y_n=\dfrac{(2k-1)!!}{(2k)!!}\,y_0\;\;(n=2k)\\[10pt] y_n=\dfrac{(2k)!!}{(2k+1)!!}\,y_  1\;\;(n=2k+1)\end{array}\right. \\[7pt]&\qquad\qquad\big(y_0=\dfrac{\pi  }{2}\,\;\;y_1=1\big)\end{aligned  }

    \bullet Lower bound:

    x\neq 0:\; e^x>x+1\Rightarrow e^{-nx^2}>(1-x^2)^n. Integrate and set x=\sin t:

    \displaystyle\begin{aligned} \int_0^{1}e^{-nx^2}dx>\int_0^1 (1-x^2)^n\,dx\Rightarrow  \int_0^{1}e^{-nx^2}dx>\int_0^{\frac{\pi}{2}} \cos^{2n+1} t\,dt=\frac{(2n)!!}{(2n+1)!!}

    Observe that \displaystyle \int_0^1 e^{-nx^2}dx<\int_0^{\infty} e^{-nx^2}\,dx=\frac{1}{\sqrt{n}}\int  _0^{\infty} e^{-x^2}\,dx and hence:

    \displaystyle  n\left(\frac{(2n)!!}{(2n+1)!!} \right)^2<\left(\int_0^{\infty} e^{-x^2}\,dx\right)^2

    \bullet Upper bound:

    Similarly, e^{x}>x+1\Rightarrow e^{-nx^2}<(x^2+1)^{-n}. This time set x=\tan t and integrate directly over the interval of interest:

    \displaystyle \begin{aligned} \int_0^{\infty}e^{-nx^2}dx<\int_0^{\infty} (x^2+1)^{-n}\,dx\Rightarrow  \int_0^{\infty}e^{-nx^2}dx<\int_0^{\frac{\pi}{2}} \cos^{2n-2} t\,dt=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}

    As above, this yields:

    \displaystyle \left(\int_0^{\infty} e^{-x^2}\,dx\right)^2< n\left(\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2} \right)^2




    Now it is well-known that \displaystyle \dfrac{2}{1}\cdot \dfrac{2}{3}\cdot \dfrac{4}{3}\cdot \dfrac{4}{5}\cdots = \lim_{n\to\infty}\frac{1}{2n+1} \left(\frac{(2n)!!}{(2n-1)!!} \right)^2=\frac{\pi}{2} hence:

    \displaystyle\begin{aligned} \frac{n}{2n+1}\left[\frac{1}{2n+1} \left(\frac{(2n)!!}{(2n-1)!!} \right)^2\right]<\left(\int_0^{\infty} e^{-x^2}\,dx\right)^2<\frac{\pi^2}{4  }\cdot\frac{n}{2n-1}\left[\frac{1}{2n-1} \left(\frac{(2n-2)!!}{(2n-3)!!} \right)^2\right]^{-1}

    \displaystyle\begin{aligned} \frac{1}{2}\left(\frac{\pi}{2} \right)\leq\left(\int_0^{\infty} e^{-x^2}\,dx\right)^2\leq\frac{\pi^2  }{4}\cdot\frac{1}{2} \left( \frac{2}{\pi}\right)\Rightarrow \left(\int_0^{\infty} e^{-x^2}\,dx\right)^2=\frac{\pi}{4}

    Thus \displaystyle\int_{-\infty}^{\infty} e^{-x^2}\,dx=\sqrt{\pi}

    Mladenov I think there is a typo in your question, the \frac{\pi}{2} should be included in the square as well.
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    (Original post by Lord of the Flies)
    Solution 130
    Well done.


    (Original post by Lord of the Flies)
    Mladenov I think there is a typo in your question, the \displaystyle \frac{\pi}{2} should be included in the square as well.
    Yes, it is a typo, I apologise.
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    (Original post by bananarama2)
    Solution 129

     \frac{d}{dx} \frac{1}{k} \int_0^{x} f(x')\sinh (k(x-x')) dx'

    = \frac{1}{k} f(x) \sinh (k(x-x)) - f(0) \frac{1}{k}\sinh(kx) \times 0 + \frac{1}{k} \int_0^x k f(x') \cosh (k(x-x')) dx'

    = \int_0^x f(x') \cosh (k(x-x')) dx'

     \frac{d}{dx} \int_0^x f(x') \cosh (k(x-x')) dx'

     = f(x)\cosh (k(x-x)) - 0 + \int_0^x kf(x')\shin (k(x-x')) dx'

     = y_1''

    SO

     y_1'' - k^2y

     = f(x)\cosh (k(x-x)) + \int_0^x kf(x')\shin (k(x-x')) dx' - \int_0^x kf(x')\shin (k(x-x')) dx'

     = f(x)
    hmmmm not sure about your multiplication by zero in the second line though you appear to get the correct answer so I'll give you that.

    For the more general case

    PROBLEM <some number>**

    Find the most general solution to the linear second order differential equation:

    y''+ay'+by=f(x)

    (there is absolutely no need to apply the wronskian btw so don't if you can help it)
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    (Original post by Lord of the Flies)
    I am not sure which was more of an adventure - doing the problem, or typing the solution.

    Solution 130

    First we prove an intermediate result:

    \displaystyle \begin{aligned}y_n=\int_0^{\pi/2}\cos^nt\,dt\Rightarrow y_n&=(n-1)\int_0^{\pi/2}\sin^2 t\cos^{n-2}t\,dt\\[7pt]&=(n-1)y_{n-2}-(n-1)y_{n}\\[7pt]&\qquad\Rightarrow y_n=\frac{n-1}{n}y_{n-2}\\[7pt]&\qquad\Rightarrow\left\{ \begin{array}{l l} y_n=\dfrac{(2k-1)!!}{(2k)!!}\,y_0\;\;(n=2k)\\[10pt] y_n=\dfrac{(2k)!!}{(2k+1)!!}\,y_  1\;\;(n=2k+1)\end{array}\right. \\[7pt]&\qquad\qquad\big(y_0=\dfrac{\pi  }{2}\,\;\;y_1=1\big)\end{aligned  }

    \bullet Lower bound:

    x\neq 0:\; e^x&gt;x+1\Rightarrow e^{-nx^2}&gt;(1-x^2)^n. Integrate and set x=\sin t:

    \displaystyle \int_0^{1}e^{-nx^2}&gt;\int_0^1 (1-x^2)^n\,dx\Rightarrow  \int_0^{1}e^{-nx^2}&gt;\int_0^{\frac{\pi}{2}} \cos^{2n+1} t\,dt=\frac{(2n)!!}{(2n+1)!!}

    Observe that \displaystyle \int_0^1 e^{-nx^2}dx&lt;\int_0^{\infty} e^{-nx^2}\,dx=\frac{1}{\sqrt{n}}\int  _0^{\infty} e^{-x^2}\,dx and hence:

    \displaystyle  n\left(\frac{(2n)!!}{(2n+1)!!} \right)^2&lt;\left(\int_0^{\infty} e^{-x^2}\,dx\right)^2

    \bullet Upper bound:

    Similarly, e^{x}&gt;x+1\Rightarrow e^{-nx^2}&lt;(x^2+1)^{-n}. This time set x=\tan t and integrate directly over the interval of interest:

    \displaystyle \int_0^{\infty}e^{-nx^2}&lt;\int_0^{\infty} (x^2+1)^{-n}\,dx\Rightarrow  \int_0^{\infty}e^{-nx^2}&lt;\int_0^{\frac{\pi}{2}} \cos^{2n-2} t\,dt=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}

    As above, this yields:

    \displaystyle \left(\int_0^{\infty} e^{-x^2}\,dx\right)^2&lt; n\left(\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2} \right)^2




    Now it is well-known that \displaystyle \dfrac{2}{1}\cdot \dfrac{2}{3}\cdot \dfrac{4}{3}\cdot \dfrac{4}{5}\cdots = \lim_{n\to\infty}\frac{1}{2n+1} \left(\frac{(2n)!!}{(2n-1)!!} \right)^2=\frac{\pi}{2} hence:

    \displaystyle\begin{aligned} \frac{n}{2n+1}\left[\frac{1}{2n+1} \left(\frac{(2n)!!}{(2n-1)!!} \right)^2\right]&lt;\left(\int_0^{\infty} e^{-x^2}\,dx\right)^2&lt;\frac{\pi^2}{4  }\cdot\frac{n}{2n-1}\left[\frac{1}{2n-1} \left(\frac{(2n-2)!!}{(2n-3)!!} \right)^2\right]^{-1}

    \displaystyle\begin{aligned} \frac{1}{2}\left(\frac{\pi}{2} \right)\leq\left(\int_0^{\infty} e^{-x^2}\,dx\right)^2\leq\frac{\pi^2  }{4}\cdot\frac{1}{2} \left( \frac{2}{\pi}\right)\Rightarrow \left(\int_0^{\infty} e^{-x^2}\,dx\right)^2=\frac{\pi}{4}

    Thus \displaystyle\int_{-\infty}^{\infty} e^{-x^2}\,dx=\sqrt{\pi}

    Mladenov I think there is a typo in your question, the \frac{\pi}{2} should be included in the square as well.
    Seeing as you appear from an earlier post to have done multiple integrals, there is a far simpler way to approach this problem
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    (Original post by und)
    Problem 1*

    Prove that if the function \displaystyle f is continuous on \displaystyle [a,b] and \displaystyle \int^b_af(x)g(x) dx = 0 for any continuous function \displaystyle g, then \displaystyle f=0 on \displaystyle [a,b].
    You can't use integration over a non-continuous function anyway though. Although you could do it with the chain rule to show you can integrate part of it.
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    (Original post by k9markiii)
    You can't use integration over a non-continuous function anyway though. Although you could do it with the chain rule to show you can integrate part of it.
    I do not know what you are trying to say, but not all integrable functions are continuous. For example, you can take any bounded function such that the set of its discontinuities has measure zero.
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    (Original post by Mladenov)
    Edit: For the sake of completeness, I would like to add that we can use Chebotarev's density theorem to show that there are many primes q which satisfy the condition. In other words, we look for those primes q for which the degree of the splitting field of P over F_{q} is p.
    What do you think about the case when \deg (P) = p^{\alpha}, where \alpha &gt; 1.
    In fact, infinitely many. If you play with Chebotarev's theorem for a while, then you will discover that such an irreducible P \in \mathbb{Z}[X], which has a root in \mathbb{Z}/q\mathbb{Z} for all but finitely many primes q, must have degree 1. An even more striking fact is that for every P \in \mathbb{Z}[X], with \deg(P) \geq 1, there is a certain number m, determined by the coefficients of the irreducible factors of P, such that P has a root in \mathbb{Z}/n\mathbb{Z} for all n \in \mathbb{N} if, and only if, it has a root in \mathbb{Z}/m\mathbb{Z} (with one more condition).

    You don't need the prime degree assumption. A monic P \in \mathbb{Z}[X], with \deg(P) &gt; 1, whose reduction in \mathbb{Z}/p\mathbb{Z}[X] has a root for all primes p must be reducible over \mathbb{Z}. This can be proved by playing with the Galois group and only a weaker form of Frobenius' density theorem.

    A good example to keep in mind, though, is x^4 + 1. This is irreducible over \mathbb{Z}, but reducible in \mathbb{Z}/p\mathbb{Z} for all primes p. The condition for having a root is stronger: the congruence x^4 \equiv -1\ (p) has no solutions for primes p \equiv 3\ (4).
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    (Original post by Lord of the Flies)
    ...and the lecturer is French!....
    Makes me think of this

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    (Original post by natninja)
    hmmmm not sure about your multiplication by zero in the second line though you appear to get the correct answer so I'll give you that.

    For the more general case

    PROBLEM <some number>**

    Find the most general solution to the linear second order differential equation:

    y''+ay'+by=f(x)

    (there is absolutely no need to apply the wronskian btw so don't if you can help it)
    Can I check the answer first? I get

     y=e^{\beta x} \int \left(e^{\alpha x - \beta x } \int f(x) e^{-\alpha x} dx \right)dx

     \alpha = \frac{-a+\sqrt{a^2-4b}}{2} \qquad \beta=  \frac{-a-\sqrt{a^2-4b}}{2}
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    (Original post by bananarama2)
    Can I check the answer first? I get

     y=e^{\beta x} \int \left(e^{\alpha x - \beta x } \int f(x) e^{-\alpha x} dx \right)dx

     \alpha = \frac{-a+\sqrt{a^2-4b}}{2} \qquad \beta=  \frac{-a-\sqrt{a^2-4b}}{2}
    the answer looks a fair bit nicer than that... but your alpha and beta are correct, your answer should be a sum of roots and you shouldn't have integrals with respect to x but with respect to a dummy variable, they should also be definite integrals
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    (Original post by natninja)
    Seeing as you appear from an earlier post to have done multiple integrals, there is a far simpler way to approach this problem
    That is a solution to problem 130

    (would you mind spoilering that quote, so that it takes up less space on the page?)

    (Original post by k9markiii)
    You can't use integration over a non-continuous function anyway though. Although you could do it with the chain rule to show you can integrate part of it.
    Yes you can.

    (Original post by ukdragon37)
    Makes me think of this
    :lol:

    I like this one too.
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    (Original post by jack.hadamard)
    An even more striking fact is that for every P \in \mathbb{Z}[X], with \deg(P) \geq 1, there is a certain number m, determined by the coefficients of the irreducible factors of P, such that P has a root in \mathbb{Z}/n\mathbb{Z} for all n \in \mathbb{N} if, and only if, it has a root in \mathbb{Z}/m\mathbb{Z} (with one more condition).
    Something more on this would be highly appreciated.

    (Original post by jack.hadamard)
    You don't need the prime degree assumption. A monic P \in \mathbb{Z}[X], with \deg(P) &gt; 1, whose reduction in \mathbb{Z}/p\mathbb{Z}[X] has a root for all primes p must be reducible over \mathbb{Z}. This can be proved by playing with the Galois group and only a weaker form of Frobenius' density theorem.
    Some difficulties arise here; but it is indeed splendid result.

    (Original post by jack.hadamard)
    In fact, infinitely many. If you play with Chebotarev's theorem for a while, then you will discover that such an irreducible P \in \mathbb{Z}[X], which has a root in \mathbb{Z}/q\mathbb{Z} for all but finitely many primes q, must have degree 1.
    Here is what I think.
    Spoiler:
    Show
    Suppose for the sake of contradiction that there are finitely many primes for which P has no root in \mathbb{Z}/q\mathbb{Z}[X], and \deg(P) &gt;1. Then, Frobenius' Density Theorem shows that each element of Gal(P) fixes a root of P. Hence, Gal(P) is the union of the conjugates of its subgroup which consists of all elements which fix a root of P. Therefore, Gal(P) is trivial, which is a contradiction.
 
 
 
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