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    (Original post by Felix Felicis)
    Another solution to 295:

    \displaystyle I = \int_{-\infty}^{\infty} e^{-x^2} \ dx = 2 \int_0^{\infty} e^{-x^2} \ dx \overset{x^2 \mapsto x}= \int_0^{\infty} e^{-x} x^{-\frac{1}{2}} \ dx = \Gamma \left(\tfrac{1}{2}\right)

    Now, as \Gamma (z) = \int_0^{\infty} x^{z-1} e^{-x} \ dx and \text{B} (x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x+y)}, we have:

    \text{B}\left(\frac{1}{2}, \frac{1}{2}\right) = \dfrac{\Gamma^2 \left(\frac{1}{2}\right)}{\Gamma (1)} = \Gamma^2 \left(\frac{1}{2}\right) = I^2

    \displaystyle \text{B}\left(\tfrac{1}{2}, \tfrac{1}{2}\right) = \int_0^1 x^{-\frac{1}{2}} (1-x)^{-\frac{1}{2}} \ dx \overset{x^{\frac{1}{2}}\mapsto x}= 2\arcsin{x}\bigg|_0^1 = \pi

    Therefore I^2 = \pi \implies I = \sqrt{\pi}
    You do love the gamma function, don't you
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    (Original post by james22)
    Is there a method of solving this integral using just elementary methods (i.e. no double integrals or change of coordinate systems).
    *fumbles to push glasses into eyeballs*.

    Consider the surface z = e^{-(x^2+y^2)} but don't actually do anything so as not to violate the terms of your post.
    The volume under this is the square of the integral we want to find.

    But this can also be thought of as the volume of revolution about the z axis where z = e^{-x^2} and so x = \sqrt{- \ln(z)}...

    Using the formula for the volume of revolution:
    \displaystyle V = \int_0^1 x^2 dx = \pi \int_0^1 - \ln(z) dz
    Just by IBP:
    \displaystyle V = - \pi [z \ln(z) - z]_{0}^{1} = \pi
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    (Original post by james22)
    Is there a method of solving this integral using just elementary methods (i.e. no double integrals or change of coordinate systems).
    Depends what you class as an elementary method. The function has no antiderivative, if that's what you mean.
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    Problem 396**

    Hopefully I'll solve Mladenov's problems at some point, but in the hiatus here's another integral:

    \displaystyle \int_0^{\frac{\pi}{2}}\frac{\log  (\sec(x))}{\tan(x)}dx.

    Problem 397**

    \displaystyle \int_0^{\frac{\pi}{2}}\frac{\log  (\sec(x))}{\tan^2(x)}dx.
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    (Original post by Mladenov)

    Problem 365***(revisited)

    Find \displaystyle \int_{0}^{\infty} \frac{x^{a}}{(b+x^{c})^{d}}dx, when \displaystyle a \ge 0, b>0, c>0, d > \frac{a+1}{c}. We also throw out the condition d \in \mathbb{Z}.
    Solution 365(revisited)***

    \displaystyle \int_{0}^{\infty} \frac{x^{a}}{(b+x^{c})^{d}}dx= \frac{1}{c} \int_{0}^{\infty} \frac{u^{\frac{a+1}{c}-1}}{(b+u)^{d}}du= \frac{b^{\frac{a+1}{c}-d}}{c} \int_{0}^{\infty} \frac{z^{\frac{a+1}{c}-1}}{(1+z)^{d}}dz

    \displaystyle =\frac{b^{\frac{a+1}{c}-d}}{c}\frac{\Gamma \left(\frac{a+1}{c}\right) \Gamma \left(d-\frac{a+1}{c}\right)}{\Gamma \left(d\right)}

    I doubt this is as simple as it gets, some result from Gauss will probably simplify the numerator further. A convenient contour may also do the trick.
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    (Original post by Flauta)
    It's still a really good question, there're some quite complex in-depth ways of doing it which I don't understand at all, would love to see some people post them. What was your method?
    I used the polar coordinates method as well

    I haven't seen any differential equations on this thread so here is one.

    Problem 398***
    Find the general solution to the system of differential equations.
     y'_1= 4y_1 - y_3
     y'_2 = 2y_1 + 2y_2 - y_3
      y'_3 = 3y_1 + y_2
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    This looks so hard, not even gonna lie
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    (Original post by techno-thriller)
    This looks so hard, not even gonna lie
    Which question?
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    (Original post by techno-thriller)
    This looks so hard, not even gonna lie
    Good! I was aiming for that first impression. :mwuaha:

    Posted from TSR Mobile
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    (Original post by techno-thriller)
    This looks so hard, not even gonna lie
    Hard is good, makes it even more satisfying when you complete a problem I'm sure there'll be something on here you could do, I manage to find the odd thing occasionally
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    (Original post by Flauta)
    Hard is good, makes it even more satisfying when you complete a problem I'm sure there'll be something on here you could do, I manage to find the odd thing occasionally
    Most of the easier questions on here are the ones that I have posted. Some would say that I have diluted the thread but in my mind, I'm making it more accessible

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    (Original post by keromedic)
    Which question?
    (Original post by Arieisit)
    Good! I was aiming for that first impression. :mwuaha:

    Posted from TSR Mobile
    (Original post by Flauta)
    Hard is good, makes it even more satisfying when you complete a problem I'm sure there'll be something on here you could do, I manage to find the odd thing occasionally
    All those squiglly lines
    Shivers me.
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    (Original post by Arieisit)
    Most of the easier questions on here are the ones that I have posted. Some would say that I have diluted the thread but in my mind, I'm making it more accessible

    Posted from TSR Mobile
    I don't understand your differential equation one that you just posted Reading a lot about differential equations at the moment though, I'm sure I'll understand it soon
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    (Original post by henpen)
    Very much so, could you quickly summarise what neutrices are? It seems counterintuitive that a function can have a defined derivative at a point where it is not defined.
    Sorry for that regrettably late reply; an elementary example of function with this property is the real logarithm.
    Neutrices were introduced, as far as I know, (I am not quite knowledgeable in the area) to help statisticians/theoretical physicists deal with asymptotic expressions, i.e. they generalize O and o notions. A neutrix is an abelian (additively written) group N, consisting of functions S \to G, where G is an abelian group, such that the only constant function in N is the zero function. The functions in N are called negligible.
    As an example consider S = \mathbb{Z}, G = \mathbb{R}, and the negligible functions in N being finite (linear) sums of n^{x} \ln^{k-1}n and \ln^{k}n, where x>0, k - positive integer, and all functions which tend to zero as n \to \infty.
    This article is relevant.
    Spoiler:
    Show
    From the theory, we have \displaystyle \beta_{(1,2)}(0,1) = - \sum_{1}^{\infty} \frac{4H_{n}}{(n+1)^{3}}.


    (Original post by henpen)
    Spoiler:
    Show
    Solution 365(revisited)***

    \displaystyle \int_{0}^{\infty} \frac{x^{a}}{(b+x^{c})^{d}}dx= \frac{1}{c} \int_{0}^{\infty} \frac{u^{\frac{a+1}{c}-1}}{(b+u)^{d}}du= \frac{b^{\frac{a+1}{c}-d}}{c} \int_{0}^{\infty} \frac{z^{\frac{a+1}{c}-1}}{(1+z)^{d}}dz

    \displaystyle =\frac{b^{\frac{a+1}{c}-d}}{c}\frac{\Gamma \left(\frac{a+1}{c}\right) \Gamma \left(d-\frac{a+1}{c}\right)}{\Gamma \left(d\right)}

    I doubt this is as simple as it gets, some result from Gauss will probably simplify the numerator further. A convenient contour may also do the trick.
    Yes, that's correct.

    After this analysis, I see that the world is beautiful only when we treat stuff algebraically.

    Problem 399***

    Let G be a group. Show that there exists a cell complex X such that \pi_{1}(X,a)=G, where a is a vertex of X. A nice application of this topological result is the existence of coproducts in the category of groups G.

    (Original post by jack.hadamard)
    If you are having a boring Wednesday night, then play with n!^{(k)} + a = m^2 to find solutions (n,m) for k,a \in \mathbb{N}.

    For example, n!! + 2 = m^2.
    The special case k=2, a=2 is easy. Clearly, if n \equiv 0 \pmod 2, then n=2. If n \equiv 1 \pmod 2, we consider our equation \pmod 3 and see that there are no solutions.
    I will do the general case tomorrow.
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    (Original post by Arieisit)
    I used the polar coordinates method as well

    I haven't seen any differential equations on this thread so here is one.

    Problem 398***
    Find the general solution to the system of differential equations.
     y'_1= 4y_1 - y_3
     y'_2 = 2y_1 + 2y_2 - y_3
      y'_3 = 3y_1 + y_2
    To be fair, since I only started learning differential just recently, I had to get some help with this question.

    A=\begin{pmatrix}4&0&-1\\2&2&-1\\3&1&0\end{pmatrix} which has a defined polynomial \chi_A=(X-2)^3. Which also has a Jordan form \begin{pmatrix}2&0&0\\1&2&0\\0&1  &2\end{pmatrix} which will give a system X'=CJC^{-1}X, which should be solved by the change of variables Y=C^{-1}X.

    But, sadly I don't know how Maybe someone better can take it from here.

    Can you please tell me where you got this question? I want to read more into these types of questions.
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    (Original post by henpen)
    Problem 396**

    Hopefully I'll solve Mladenov's problems at some point, but in the hiatus here's another integral:

    \displaystyle \int_0^{\frac{\pi}{2}}\frac{\log  (\sec(x))}{\tan(x)}dx.

    Problem 397**

    \displaystyle \int_0^{\frac{\pi}{2}}\frac{\log  (\sec(x))}{\tan^2(x)}dx.
    Solution 397
    Seemed a bit too simple so most probably I made mistake (I hope log(x)=ln(x))
    u=tanx
    \displaystyle \int \frac{\log(\sec(x))}{\tan^2(x)}d  x=\dfrac{1}{2} \int \dfrac{log(u^2+1)}{u^2}du \  _{\rightarrow}^{IBP} \ \ \ \ \dfrac{1}{2} \left( 2arctanu-\dfrac{log(u^2+1)}{u} \right)
    \therefore \displaystyle \int_0^{\frac{\pi}{2}}\frac{\log  (\sec(x))}{\tan(x)}dx= \dfrac{1}{2} \left[arctan(tan(x))-\dfrac{sec^2x}{tanx} \right]_0^{\frac{\pi}{2}}
    Assuming I've made no mistakes (surprising as it's midnight), I'm stuck as I'm not sure how to evaluate arctan(tan(\frac{\pi}{2})) although googling suggests it might be 0. Not sure though.
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    (Original post by keromedic)
    Solution 397
    Seemed a bit too simple so most probably I made mistake (I hope log(x)=ln(x))
    u=tanx
    \displaystyle \int \frac{\log(\sec(x))}{\tan^2(x)}d  x=\dfrac{1}{2} \int \dfrac{log(u^2+1)}{u^2}du \  _{\rightarrow}^{IBP} \ \ \ \ \dfrac{1}{2} \left( 2arctanu-\dfrac{log(u^2+1)}{u} \right)
    \therefore \displaystyle \int_0^{\frac{\pi}{2}}\frac{\log  (\sec(x))}{\tan(x)}dx= \dfrac{1}{2} \left[arctan(tan(x))-\dfrac{sec^2x}{tanx} \right]_0^{\frac{\pi}{2}}
    Assuming I've made no mistakes (surprising as it's midnight), I'm stuck as I'm not sure how to evaluate arctan(tan(\frac{\pi}{2})) although googling suggests it might be 0. Not sure though.
    Well, \arctan{\tan{x}} = x as you're integrating over \left[0,\frac{\pi}{2}\right] but you've done the substitution incorrectly, you've forgotten to make an appropriate substitution for dx. Your integral is actually transformed to:

    \displaystyle \int_0^{\frac{\pi}{2}} \frac{\ln{\sec{x}}}{\tan^2 x} \ dx \overset{u=\tan{x}}= \frac{1}{2}\int_0^{\infty} \frac{\ln{(u^2 + 1)}}{u^2 (u^2 + 1)} \ du.
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    (Original post by Felix Felicis)
    Well, \arctan{\tan{x}} = x as you're integrating over \left[0,\frac{\pi}{2}\right] but you've done the substitution incorrectly, you've forgotten to make an appropriate substitution for dx. Your integral is actually transformed to:

    \displaystyle \int_0^{\frac{\pi}{2}} \frac{\ln{\sec{x}}}{\tan^2 x} \ dx \overset{u=\tan{x}}= \frac{1}{2}\int_0^{\infty} \frac{\ln{(u^2 + 1)}}{u^2 (u^2 + 1)} \ du.
    oH. I can't believe I did that. sorry guys
    :facepalm:
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    (Original post by Flauta)
    Depends what you class as an elementary method. The function has no antiderivative, if that's what you mean.
    I know, I mean something working with just the basic methods of integration (like substitution and IBP etc).
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    (Original post by MAyman12)
    To be fair, since I only started learning differential just recently, I had to get some help with this question.

    A=\begin{pmatrix}4&0&-1\\2&2&-1\\3&1&0\end{pmatrix} which has a defined polynomial \chi_A=(X-2)^3. Which also has a Jordan form \begin{pmatrix}2&0&0\\1&2&0\\0&1  &2\end{pmatrix} which will give a system X'=CJC^{-1}X, which should be solved by the change of variables Y=C^{-1}X.

    But, sadly I don't know how Maybe someone better can take it from here.

    Can you please tell me where you got this question? I want to read more into these types of questions.

    Yes, so far so good but remember that so far you have only attempted to solve the first single 1-dimensional differential equation.

    I stumbled on Systems of Differential equations on a maths website that I was reading because obviously I didn't learn this in A levels :rolleyes:

    The question is from a pdf file so I can't upload it to this thread since it has the solution as well.
 
 
 
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