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    (Original post by ζ(s))
    \begin{aligned} \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{x} \, dx & = \int_{0}^{1} \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{n+1}\,dx =  \sum_{n=0}^{\infty}\int_{0}^{1} \frac{(-1)^nx^{2n+1}}{n+1}\,dx \\& = \sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)(2n+2)} = \frac{1}{2} \sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^2}  \\& = \frac{1}{2} \left(1-\frac{1}{2}\right)\sum_{n=0}^{ \infty} \frac{1}{(n+1)^2} = \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} \\& = \frac{1}{4}\left(\frac{\pi^2}{6} \right) = \frac{\pi^2}{24}. \end{aligned}
    beautiful username bernhard
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    Not at all a contribution to this thread, and I know the title is obviously tongue in cheek since most of the stuff in here is not trivial but...


    I absolutely think the sentence "the proof is trivial" should ABSOLUTELY BE BANNED from university lecture notes.

    I cannot stand it when you open a set of lecture notes on something you understand precisely nothing about, and within the first couple of pages the author is stating theorem after theorem "leaving the proof of the theorem to the reader" because "the proof is trivial" . Yes it may be trivial IF you understand the topic but if I understood said topic I wouldn't be reading the first couple of pages on a set of notes for it...

    /Rant over .
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    (Original post by ζ(s))
    \begin{aligned} \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{x} \, dx & = \int_{0}^{1} \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{n+1}\,dx =  \sum_{n=0}^{\infty}\int_{0}^{1} \frac{(-1)^nx^{2n+1}}{n+1}\,dx \\& = \sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)(2n+2)} = \frac{1}{2} \sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^2}  \\& = \frac{1}{2} \left(1-\frac{1}{2}\right)\sum_{n=0}^{ \infty} \frac{1}{(n+1)^2} = \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} \\& = \frac{1}{4}\left(\frac{\pi^2}{6} \right) = \frac{\pi^2}{24}. \end{aligned}
    This is a thing of beauty. I would have just contour-integrated, and my answer would have been horrible.
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    Now how about

    Problem 467**

     \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{1 + x} \, dx ?

    though perhaps the way I did it was over-complicated, I don't know
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    (Original post by jjpneed1)
    Now how about

    Problem 467**

     \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{1 + x} \, dx ?

    though perhaps the way I did it was over-complicated, I don't know
    Solution 467

    \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{1 + x} dx



=\displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{1} dx + \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{x} dx


    = \frac{\pi^2}{24}+-2+\frac{\pi}{2}+log(2)

    Using the previous result and integrating log(1+x^2) by parts (basic integral, nothing worth showing).

    EDIT: I've just realised this is a load of rubbish.
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    (Original post by james22)
    Solution 467

    \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{1 + x} dx



=\displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{1} dx + \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{x} dx


    = \frac{\pi^2}{24}+-2+\frac{\pi}{2}+log(2)

    Using the previous result and integrating log(1+x^2) by parts (basic integral, nothing worth showing).
    How did you get from the first line to the second line?
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    (Original post by majmuh24)
    Can you break up a fraction from the sum in the denominator like that? I thought that only worked in the numerator

    Posted from TSR Mobile
    You can't (usually) so it's either an error or a special case in this situation.
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    (Original post by james22)
    Solution 467

    \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{1 + x} dx



=\displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{1} dx + \displaystyle \int_0^{1} \frac{\ln(1 + x^2)}{x} dx


    = \frac{\pi^2}{24}+-2+\frac{\pi}{2}+log(2)

    Using the previous result and integrating log(1+x^2) by parts (basic integral, nothing worth showing).
    Not correct I'm afraid, I've never seen an integral broken up like that before but I guess you had a reason?
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    (Original post by jjpneed1)
    Not correct I'm afraid, I've never seen an integral broken up like that before but I guess you had a reason?
    Did you use integration by parts?

    Spoiler:
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    I've used the substitutions x=\tan \theta, then u=1+\tan \theta and I've ended up at:

    \begin{aligned} \displaystyle \int_{0}^{1} \dfrac{\log(1+x^2)}{1+x} \text{ d}x & \ \overset{x=\tan\theta}= \int_{0}^{\frac{\pi}{4}} \dfrac{\log(1+\tan^2 \theta) \sec^2 \theta}{1+\tan \theta} \text{ d}\theta \\ & \overset{u=1+\tan\theta}= \int_{1}^{2} \dfrac{\log(u^2 - 2u + 2)}{u} \text{ d}u \end{aligned}

    Edit: I'm going to try and express everything in terms of \sin \theta and \cos \theta after the first substitution.

    Edit2: Nope, nothing.
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    I don't like integrals. But you guys sure do. Here's one I can only do via complex analysis; I wonder if anyone can manage it without it:

    Problem 468***

    \displaystyle\int_0^{\pi}\dfrac{  \cos^2 x}{5+4\cos x}\ dx
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    (Original post by FireGarden)
    I don't like integrals. But you guys sure do. Here's one I can only do via complex analysis; I wonder if anyone can manage it without it:

    Problem 468***

    \displaystyle\int_0^{\pi}\dfrac{  \cos^2 x}{5+4\cos x}\ dx
    You're joking right?

    Spoiler:
    Show

    (Original post by Khallil)
    Did you use integration by parts?

    Spoiler:
    Show
    I've used the substitutions x=\tan \theta, then u=1+\tan \theta and I've ended up at:

    \begin{aligned} \displaystyle \int_{0}^{1} \dfrac{\log(1+x^2)}{1+x} \text{ d}x & \ \overset{x=\tan\theta}= \int_{0}^{\frac{\pi}{4}} \dfrac{\log(1+\tan^2 \theta) \sec^2 \theta}{1+\tan \theta} \text{ d}\theta \\ & \overset{u=1+\tan\theta}= \int_{1}^{2} \dfrac{\log(u^2 - 2u + 2)}{u} \text{ d}u \end{aligned}

    Edit: I'm going to try and express everything in terms of \sin \theta and \cos \theta after the first substitution.

    Edit2: Nope, nothing.
    In case you're interested in how I did it, here's a hint:
    Spoiler:
    Show
    differentiation under the integral sign

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    (Original post by jjpneed1)
    You're joking right?
    Why would I be joking? Have you seen half the integrals in this thread.. I just wolfram'd this thing and it has one gnarly antiderivative, but even then, some integrals in this thread don't even have antiderivatives at all.
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    (Original post by FireGarden)
    Why would I be joking? Have you seen half the integrals in this thread.. I just wolfram'd this thing and it has one gnarly antiderivative, but even then, some integrals in this thread don't even have antiderivatives at all.
    Surely you would try long division before complex analysis?? Then it's just a t-sub away :confused:
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    (Original post by Khallil)
    How did you get from the first line to the second line?

    (Original post by majmuh24)
    Can you break up a fraction from the sum in the denominator like that? I thought that only worked in the numerator

    Posted from TSR Mobile

    (Original post by keromedic)
    You can't (usually) so it's either an error or a special case in this situation.

    (Original post by jjpneed1)
    Not correct I'm afraid, I've never seen an integral broken up like that before but I guess you had a reason?
    I din't do anything special, my brain just stopped functioning.
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    (Original post by jjpneed1)
    Surely you would try long division before complex analysis?? Then it's just a t-sub away :confused:
    Well, I did say I don't really like integrals! If a sledge hammer I have I know will work and I have no other immediate ideas, I usually get on with it. Have you tried it, though?
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    (Original post by FireGarden)
    Well, I did say I don't really like integrals! If a sledge hammer I have I know will work and I have no other immediate ideas, I usually get on with it. Have you tried it, though?
    Oh fair enough, I apologise! Yeah I did try it and got  5\pi/24
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    (Original post by jjpneed1)
    Oh fair enough . Yeah I did try it and got  5\pi/24
    That's what I got. Indeed, my integration ability is embarrassing
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    (Original post by FireGarden)
    That's what I got. Indeed, my integration ability is embarrassing
    No way, if I knew complex analysis I'd try and use it on every problem, it's pretty cool
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    (Original post by jjpneed1)
    No way, if I knew complex analysis I'd try and use it on every problem, it's pretty cool
    Nah, DUTIS is way cooler. It seems like absolute magic.
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    (Original post by Smaug123)
    Nah, DUTIS is way cooler. It seems like absolute magic.
    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}}
 
 
 
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