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    (Original post by FireGarden)
    If cool is what we want, here's the coolest evaluation of an integral I've seen.

    We want to evaluate \displaystyle\int_{-\infty}^{\infty} e^{-ax^2}\ dx = f(a).

    We will proceed with dimensional analysis. Suppose x represents some length, so it has dimension [L]. The exp function is required to have dimensionless input (as it is dimensionless itself), so a must have dimension [L]^{-2}, and dx of course has dimension [L].


    Overall the LHS has dimension [1][L]. Thus f(a) has these dimensions too, so f(a)=\frac{k}{\sqrt{a}} where k is a dimensionless constant. In particular, f(1)=\sqrt{\pi} which defines this constant, hence f(a)=\sqrt{\frac{\pi}{a}}
    Ah, I've had this hinted at by Siklos, but he didn't actually give an example - that is really really neat

    EDIT: it's actually nicer than the DUTIS way, I think!
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    (Original post by FireGarden)
    If cool is what we want, here's the coolest evaluation of an integral I've seen.

    We want to evaluate \displaystyle\int_{-\infty}^{\infty} e^{-ax^2}\ dx = f(a).

    We will proceed with dimensional analysis. Suppose x represents some length, so it has dimension [L]. The exp function is required to have dimensionless input (as it is dimensionless itself), so a must have dimension [L]^{-2}, and dx of course has dimension [L].


    Overall the LHS has dimension [1][L]. Thus f(a) has these dimensions too, so f(a)=\frac{k}{\sqrt{a}} where k is a dimensionless constant. In particular, f(1)=\sqrt{\pi} which defines this constant, hence f(a)=\sqrt{\frac{\pi}{a}}
    that is just beautiful. I knew you could use dimensional analysis to check forms of equations and build them up, but to actually get f(a) like that is wow!

    I think you could use a similar idea for the arctanx integral then.
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    (Original post by FireGarden)
    If cool is what we want, here's the coolest evaluation of an integral I've seen.

    We want to evaluate \displaystyle\int_{-\infty}^{\infty} e^{-ax^2}\ dx = f(a).

    We will proceed with dimensional analysis. Suppose x represents some length, so it has dimension [L]. The exp function is required to have dimensionless input (as it is dimensionless itself), so a must have dimension [L]^{-2}, and dx of course has dimension [L].


    Overall the LHS has dimension [1][L]. Thus f(a) has these dimensions too, so f(a)=\frac{k}{\sqrt{a}} where k is a dimensionless constant. In particular, f(1)=\sqrt{\pi} which defines this constant, hence f(a)=\sqrt{\frac{\pi}{a}}
    That very nice, but can it be made rigorous?
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    (Original post by james22)
    That very nice, but can it be made rigorous?
    You can DUTIS to check the answer, I suppose, but that's no easier than the original :P
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    (Original post by james22)
    That very nice, but can it be made rigorous?
    Well, I'm no expert but it is rigourous. Another way to look at it is to take the 'units' as being some factor to do substitutions with. In the original integral, you can sub x\mapsto kx, a\mapsto\frac{a}{k^2}, then you'll get the functional equation f(a)=\dfrac{1}{k}f(\dfrac{a}{k^2  }) , which I guess* leads to the unique solution f(a)=\dfrac{f(1)}{\sqrt{a}}.

    *I guess comes from the initial note - I'm no expert! I don't know much about functional equations at all, but it will surely have some theory that shows this is entirely rigourous.
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    (Original post by FireGarden)
    If cool is what we want, here's the coolest evaluation of an integral I've seen.

    We want to evaluate \displaystyle\int_{-\infty}^{\infty} e^{-ax^2}\ dx = f(a).

    We will proceed with dimensional analysis. Suppose x represents some length, so it has dimension [L]. The exp function is required to have dimensionless input (as it is dimensionless itself), so a must have dimension [L]^{-2}, and dx of course has dimension [L].


    Overall the LHS has dimension [1][L]. Thus f(a) has these dimensions too, so f(a)=\frac{k}{\sqrt{a}} where k is a dimensionless constant. In particular, f(1)=\sqrt{\pi} which defines this constant, hence f(a)=\sqrt{\frac{\pi}{a}}
    :zomg: I never thought Dimensional Analysis could be used like that.
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    (Original post by jjpneed1)
    In case you're interested in how I did it, here's a hint:
    Spoiler:
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    differentiation under the integral sign
    Could you post this calculation, please? I can't see it. :no:
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    (Original post by ζ(s))
    Could you post this calculation, please? I can't see it. :no:
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    Consider  \displaystyle I(a) = \frac{ln(1 + ax^2)}{(1 + x)} then differentiate under the integral with respect to a. It should be clear what to do from there, just some arctan and simpler natural log integrals if I recall. If you want I can send my solution.
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    (Original post by jjpneed1)
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    Consider  \displaystyle I(a) = \frac{ln(1 + ax^2)}{(1 + x)} then differentiate under the integral with respect to a. It should be clear what to do from there, just some arctan and simpler natural log integrals if I recall. If you want I can send my solution.
    I don't think the works because, if I recall, one of the integrals that you end up getting isn't simple at all.
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    (Original post by ζ(s))
    I don't think the works because, if I recall, one of the integrals that you end up getting isn't simple at all.
    It certainly works. I just checked my working and the only non-simple integral is in fact the integral in the last problem which you have to integrate indefinitely this time. So just give a series solution for it (essentially the last problem, but without plugging in limits) which should make it the integral requiring the least amount of work
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    (Original post by jjpneed1)
    It certainly works. I just checked my working and the only non-simple integral is in fact the integral in the last problem which you have to integrate indefinitely this time. So just give a series solution for it (essentially the last problem, but without plugging in limits) which should make it the integral requiring the least amount of work
    Could you post the solution, please, if you don't mind? I'm not sure how you have done it with the series.
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    (Original post by ζ(s))
    Could you post the solution, please, if you don't mind? I'm not sure how you have done it with the series.
    Spoiler:
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     \displaystyle I(\alpha) = \int_0^1 \frac{\ln(1 + \alpha x^2)}{1 + x} \text{ d}x hence
     \displaystyle \begin{aligned} I'(\alpha) = \int_0^1 \frac{x^2}{(1+\alpha x^2)(1 + x)} \text{ d}x = \int_0^1 \frac{x}{1 + \alpha x^2} \text{ d}x - \int_0^1 \frac{1}{1+ \alpha x^2} \text{ d}x + \int_0^1 \frac{1}{(1 + x)(1 + \alpha x^2)} \text{ d}x

& = \displaystyle \frac{\ln(1 + \alpha)}{2\alpha} - \frac{\arctan(\sqrt{\alpha})}{\s  \sqrt{\alpha}} + \frac{\sqrt{\alpha}\arctan(\sqrt  {\alpha})}{1 + \alpha} - \frac{\ln(1 + \alpha)}{2(1 + \alpha)} + \frac{\ln2}{1 + \alpha}

    Integrating term by term -

     \displaystyle \begin{aligned} \frac{1}{2}\int \frac{\ln(1 + \alpha)}{\alpha} \text{ d}\alpha = \frac{1}{2}\int \sum_{n=1}^{\infty}\frac{(-1)^n^+^1\alpha^n^-^1}{n} \text{d}\alpha = \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^n^+^1}{n}\int \alpha^n^-^1 \text{ d}\alpha = \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^n^+^1\alpha^n}{n^2} \text{d}\alpha

     \displaystyle \begin{aligned} \int \frac{1}{\sqrt{\alpha}} \arctan(\sqrt{\alpha}) \text{d}\alpha = 2\sqrt{\alpha} \arctan(\sqrt{\alpha}) - \ln(1 + \alpha) + C_2 \text{(IBP)}

     \displaystyle \begin{aligned} \int \frac{\sqrt{\alpha} \arctan(\sqrt{\alpha})}{1 + \alpha} \text{d}\alpha = 2\sqrt{\alpha} \arctan(\sqrt{\alpha}) - \arctan^2(\sqrt{\alpha}) - \ln(1 + \alpha) + C_3 \text{(IBP)}

    Note that after IBP we are left with  \displaystyle \begin{aligned} \int \frac{\arctan(\sqrt{\alpha})}{\s  \sqrt{\alpha}(1 + \alpha)} \text{d}\alpha which is easily dealt with using  \displaystyle \alpha \to tan^2\theta

     \displaystyle \begin{aligned} \frac{1}{2} \int \frac{\ln(1 + \alpha)}{1 + \alpha} \text{d}\alpha = \frac{1}{4} \ln^2(1 + \alpha) + C_4 \text{(IBP)}

     \displaystyle \begin{aligned} \ln2 \int \frac{1}{1 + \alpha} \text{d}\alpha = \ln2\ln(1 + \alpha) + C_5

    Combining these results in the appropriate way we heroically obtain

     \displaystyle \begin{aligned} \int_0^1 \frac{\ln(1 + \alpha x^2)}{1 + x} \text{d}\alpha = \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^n^+^1}{n^2}\alpha^n + \ln2\ln(1 + \alpha) - \frac{1}{4} \ln^2(1 + \alpha) - \arctan^2(\sqrt{\alpha}) + C_T .

    Observe that  \displaystyle I(0) = 0 \implies C_T = 0 hence  \displaystyle I(1) = \frac{3}{4} \ln^22 - \frac{\pi^2}{48}


    Apologies for any errors that may arise.. took me quite a while to do
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    (Original post by jjpneed1)
    ...
    Brilliant stuff! It never dawned on me to ignore the non-closed form and that it would disappear in the end due to the initial condition. Even when you spelled it out as a hint, I was like 'what sorcery is this?' Thanks for typing this up.
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    Problem 9001:

    Prove that:

    \displaystyle \int_0^\infty \frac {\ln x} {(1+x^2)^2}\,\mathrm d x = - \frac {\pi} 4

    Edit: Made it a 'show that' question instead.
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    (Original post by Tarquin Digby)
    Problem 9001:

    Prove that:

    \displaystyle \int_0^\infty \frac {\ln x} {(1+x^2)^2}\,\mathrm d x = - \frac {\pi} 4

    Edit: Made it a 'show that' question instead.
    Spoiler:
    Show


    \displaystyle I = \int_{0}^{\infty} \frac{\ln{x}}{(1+x^2)^2}\;{dx} = \int_{0}^{1} \frac{\ln{x}}{(1+x^2)^2}\;{dx}+ \int_{1}^{\infty} \frac{\ln{x}}{(1+x^2)^2}\;{dx}

    Let x \mapsto \frac{1}{x} then \displaystyle \int_{1}^{\infty} \frac{\ln{x}}{(1+x^2)^2}\;{dx} =- \int_{0}^{1} \frac{x^2\ln{x}}{(1+x^2)^2}\;{dx  }, thus:

    \displaystyle \int_{0}^{\infty} \frac{\ln{x}}{(1+x^2)^2}\;{dx} = \int_{0}^{1} \frac{\ln{x}}{(1+x^2)^2}\;{dx}- \int_{0}^{1} \frac{x^2\ln{x}}{(1+x^2)^2}\;{dx  }


    From here on we will need two quite simple results to unlock the integral.

    Result/Lemma #1: \displaystyle \frac{1}{(1+x^2)^2} = \sum_{k \ge 0} (-1)^k(k+1)x^{2k}.

    Proof: The usual ways (differentiating the geometric series being the easiest).

    Result/Lemma #2: \displaystyle \frac{1}{(1+n)^{2}} = - \int_{0}^{1}x^{n}\ln{x}\;{dx}.

    Proof: Your favourite method (e.g. by parts, among other methods).

    Applying result/lemma #1 first and then result/lemma #2, we have the following:

    Spoiler:
    Show
    \begin{aligned} \displaystyle I & = \int_{0}^{1} \sum_{k \ge 0}\ln{x}~(-1)^k(k+1)x^{2k}\;{dx}-\int_{0}^{1} \sum_{k \ge 0}\ln{x}~(-1)^k(k+1)x^{2k+2}\;{dx} \\& =  \sum_{k \ge 0}  \int_{0}^{1} \ln{x}~ (-1)^k(k+1)x^{2k}\;{dx}-\sum_{k \ge 0} \int_{0}^{1} \ln{x} ~(-1)^k(k+1)x^{2k+2}\;{dx} \\& = \sum_{ k \ge 0} \frac{(-1)^k (k+1)}{(2k+3)^2} -  \sum_{ k \ge 0} \frac{(-1)^k (k+1)}{(2k+1)^2} \end{aligned}

    Now we write it as an integral again, and use result/lemma #1 again!


    Spoiler:
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    \begin{aligned} I &  = \sum_{ k \ge 0}\int_{0}^{1}\int_{0}^{1} (-1)^k (k+1)  x^{2k+2}\;{dx}\;{dy} -\sum_{ k \ge 0}\int_{0}^{1}\int_{0}^{1} (-1)^k (k+1)  x^{2k}\;{dx}\;{dy}\\& =\int_{0}^{1}\int_{0}^{1}\sum_{ k \ge 0}(-1)^k (k+1)  x^{2k+2}\;{dx}\;{dy} -\int_{0}^{1}\int_{0}^{1} \sum_{ k \ge 0} (-1)^k (k+1)  x^{2k}\;{dx}\;{dy} \\& =  \int_{0}^{1} \int_{0}^{1} \frac{x^2y^2}{(1+x^2y^2)^2}\;{dx  }\;{dy}-\int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2y^2)^2}\;{dx}\;{d  y} \\& = \frac{1}{2} \int_{0}^{1} \left(\frac{\tan^{-1}{y}}{y}-\frac{1}{1+y^2}\right)\;{dy}-\frac{1}{2} \int_{0}^{1} \left(\frac{\tan^{-1}{y}}{y}+\frac{1}{1+y^2}\right)  \;{dy} \\&= -\frac{1}{2} \int_{0}^{1} \frac{1}{1+y^2}\;{dy}-\frac{1}{2}\int_{0}^{1} \frac{1}{1+y^2}\;{dy} = -\int_{0}^{1} \frac{1}{1+y^2}\;{dy} = -\frac{\pi}{4}.  \end{aligned}
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    (Original post by ζ(s))
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    \displaystyle I = \int_{0}^{\infty} \frac{\ln{x}}{(1+x^2)^2}\;{dx} = \int_{0}^{1} \frac{\ln{x}}{(1+x^2)^2}\;{dx}+ \int_{1}^{\infty} \frac{\ln{x}}{(1+x^2)^2}\;{dx}

    Let x \mapsto \frac{1}{x} then \displaystyle \int_{1}^{\infty} \frac{\ln{x}}{(1+x^2)^2}\;{dx} =- \int_{0}^{1} \frac{x^2\ln{x}}{(1+x^2)^2}\;{dx  }, thus:

    \displaystyle \int_{0}^{\infty} \frac{\ln{x}}{(1+x^2)^2}\;{dx} = \int_{0}^{1} \frac{\ln{x}}{(1+x^2)^2}\;{dx}- \int_{0}^{1} \frac{x^2\ln{x}}{(1+x^2)^2}\;{dx  }


    From here on we will need two quite simple results to unlock the integral.

    Fact/Lemma #1: \displaystyle \frac{1}{(1+x^2)^2} = \sum_{k \ge 0} (-1)^k(k+1)x^{2k}.

    Proof: The usual ways (differentiating the geometric series being the easiest).

    Fact/Lemma #2: \displaystyle \frac{1}{(1+n)^{2}} = - \int_{0}^{1}x^{n}\ln{x}\;{dx}.

    Proof: Your favourite method (e.g. by parts, among other methods).

    Applying fact/lemma #1 first and then fact/lemma #2, we have the following:

    \begin{aligned} \displaystyle I & = \int_{0}^{1} \sum_{k \ge 0}\ln{x}~(-1)^k(k+1)x^{2k}\;{dx}-\int_{0}^{1} \sum_{k \ge 0}\ln{x}~(-1)^k(k+1)x^{2k+2}\;{dx} \\& =  \sum_{k \ge 0}  \int_{0}^{1} \ln{x}~ (-1)^k(k+1)x^{2k}\;{dx}-\sum_{k \ge 0} \int_{0}^{1} \ln{x} ~(-1)^k(k+1)x^{2k+2}\;{dx} \\& = \sum_{ k \ge 0} \frac{(-1)^k (k+1)}{(2k+3)^2} -  \sum_{ k \ge 0} \frac{(-1)^k (k+1)}{(2k+1)^2} \end{aligned}

    Now we write it as an integral again, and use fact/lemma #1 again!


    \begin{aligned} I &  = \sum_{ k \ge 0}\int_{0}^{1}\int_{0}^{1} (-1)^k (k+1)  x^{2k+2}\;{dx}\;{dy} -\sum_{ k \ge 0}\int_{0}^{1}\int_{0}^{1} (-1)^k (k+1)  x^{2k}\;{dx}\;{dy}\\& =\int_{0}^{1}\int_{0}^{1}\sum_{ k \ge 0}(-1)^k (k+1)  x^{2k+2}\;{dx}\;{dy} -\int_{0}^{1}\int_{0}^{1} \sum_{ k \ge 0} (-1)^k (k+1)  x^{2k}\;{dx}\;{dy} \\& =  \int_{0}^{1} \int_{0}^{1} \frac{x^2y^2}{(1+x^2y^2)^2}\;{dx  }\;{dy}-\int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2y^2)^2}\;{dx}\;{d  y} \\& = \frac{1}{2} \int_{0}^{1} \left(\frac{\tan^{-1}{y}}{y}-\frac{1}{1+y^2}\right)\;{dy}-\frac{1}{2} \int_{0}^{1} \left(\frac{\tan^{-1}{y}}{y}+\frac{1}{1+y^2}\right)  \;{dy} \\&= -\frac{1}{2} \int_{0}^{1} \frac{1}{1+y^2}\;{dy}-\frac{1}{2}\int_{0}^{1} \frac{1}{1+y^2}\;{dy} = -\int_{0}^{1} \frac{1}{1+y^2}\;{dy} = -\frac{\pi}{4}.  \end{aligned}
    Spoiler:
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    O.O

    Alternatively use integration by parts on  \displaystyle \int_0^1 \frac{(1 - x^2)\ln(x)}{(1 + x^2)^2} {dx} and the result is immediate. Though your way looks cooler I admit
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    (Original post by jjpneed1)
    O.O

    Alternatively use integration by parts on  \displaystyle \int_0^1 \frac{(1 - x^2)\ln(x)}{(1 + x^2)^2} {dx} and the result is immediate. Though your way looks cooler I admit
    If we want cool, how about defining \displaystyle I(\alpha) = \int_{0}^{\infty}\frac{\ln{x}}{ \alpha^2+x^2}\;{dx} then putting  \displaystyle x \mapsto \alpha x.
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    (Original post by ζ(s))
    If we want cool, how about defining \displaystyle I(\alpha) = \int_{0}^{\infty}\frac{\ln{x}}{ \alpha^2+x^2}\;{dx} then putting  \displaystyle x \mapsto \alpha x.
    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
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    Why are the problems up to the 9000s..?
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    (Original post by FireGarden)
    In particular, f(1)=\sqrt{\pi} which defines this constant
    Where is this coming from? I don't see how you know this unless you know how to evaluate the integral with a=1 in the usual fashion.
 
 
 
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