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    The solution I have only uses A level knowledge. But it requires a rather insane observation. I would discount it as a possibility, to be honest!
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    (Original post by FireGarden)
    The solution I have only uses A level knowledge. But it requires a rather insane observation. I would discount it as a possibility, to be honest!
    I always love a challenge
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    (Original post by FireGarden)
    Now I'm going to be a **** head and post something I think not even LotF will evaluate without a lot of head scratching. The solution I have is only using knowledge from A level.. but I doubt it'll be used. It has a nice answer, too.

    Problem 266***

    Evaluate  \displaystyle\int_{0}^{\frac{\pi  }{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}} \ dx
     \displaystyle\int_{0}^{\frac{\pi  }{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}} \ dx =\int_{0}^{\frac{\pi}{2}} \dfrac{ \tan(x)^{\sqrt{2}}}{1+ \tan(x)^{\sqrt{2}}} \ dx\;\;(x\to \tfrac{\pi}{2}-x)

    Add both, result is \frac{\pi}{4}
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    ...That's just ridiculous. And outstandingly beautiful. I shall be back.
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    (Original post by Lord of the Flies)
     \displaystyle\int_{0}^{\frac{\pi  }{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}} \ dx =\int_{0}^{\frac{\pi}{2}} \dfrac{ \tan(x)^{\sqrt{2}}}{1+ \tan(x)^{\sqrt{2}}} \ dx\;\;(x\to \tfrac{\pi}{2}-x)

    Add both, result is \frac{\pi}{4}
    Is there anything you can't do?
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    (Original post by MathsNerd1)
    Is there anything you can't do?
    It's annoying isn't it
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    (Original post by bananarama2)
    It's annoying isn't it
    I thought after learning a few more tricks I'd be able to answer these questions, sadly I'm not of this calibre
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    (Original post by FireGarden)
    Now I'm going to be a **** head and post something I think not even LotF will evaluate without a lot of head scratching. The solution I have is only using knowledge from A level.. but I doubt it'll be used. It has a nice answer, too.

    Problem 266***

    Evaluate  \displaystyle\int_{0}^{\frac{\pi  }{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}} \ dx
    Is it pi/4? I want to check it's right before posting my working.

    EDIT: Dam, beaten to it. I used the same method as LOTF. Is there a way to do it with differentiation under the integral?
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    (Original post by MathsNerd1)
    I thought after learning a few more tricks I'd be able to answer these questions, sadly I'm not of this calibre
    Makes me wish I'd seen the light at a younger age.

    To be fair he posted this trick earlier
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    (Original post by bananarama2)
    Makes me wish I'd seen the light at a younger age.

    To be fair he posted this trick earlier
    Yeah, I only focused at the start of this year really, far too late really and yeah I've seen it a few times before but just forgot about it when first looking at the integral, oh well I'd love a good Laplace transform question as I've recently covered them
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    Next try. Taking an integral of not too much difficulty, and making it ugly. Though the "ugliness" can be thrown out the window, given the use of the right theorem, which i believe is not widely known. Let's see

    Problem 267**

    Evaluate  \displaystyle\int_{0}^{\pi} \dfrac{x\sin(x)}{1+\cos^2(x)} \ dx
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    Solution 267

    \displaystyle\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx=\int_0^{\pi} \frac{(\pi-x)\sin x}{1+\cos^2 x}dx\quad(x\to \pi-x)

    Hence I=\displaystyle\frac{\pi}{2}\int  _0^{\pi}\frac{\sin x}{1+\cos^2 x}\,dx=\frac{\pi}{2}\int_{-1}^{1}\frac{dt}{1+t^2}=\frac{\pi  ^2}{4}
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    (Original post by james22)
    Is there a way to do it with differentiation under the integral?
    Sure.

    f(s)=\displaystyle\int_0^{\frac{  \pi}{2}} \frac{dx}{1+\tan^s x}\Rightarrow f'(s)=\int_0^{\frac{\pi}{2}} -\frac{\ln  \tan x \tan^s x}{(1+\tan^s x)^2}=0

    (let x\mapsto x+\frac{\pi}{4}, use \tan (x+\frac{\pi}{4})=\cot (\frac{\pi}{4}-x) to show that the integrand is odd)

    Hence f is constant, set s=0 and we have \displaystyle f(s)=\int_0^{\frac{\pi}{2}} \frac{dx}{2}=\frac{\pi}{4}
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    Exactly how much have you practiced integration? It's just silly how quickly you do basically all of them!
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    (Original post by Lord of the Flies)
    Sure.

    f(s)=\displaystyle\int_0^{\frac{  \pi}{2}} \frac{dx}{1+\tan^s x}\Rightarrow f'(s)=\int_0^{\frac{\pi}{2}} -\frac{\ln  \tan x \tan^s x}{(1+\tan^s x)^2}=0

    (let x\mapsto x+\frac{\pi}{4}, use \tan (x+\frac{\pi}{4})=\cot (\frac{\pi}{4}-x) to show that the integrand is odd)

    Hence f is constant, set s=0 and we have \displaystyle f(s)=\int_0^{\frac{\pi}{2}} \frac{dx}{2}=\frac{\pi}{4}
    There's a little puzzle that's been niggling at me - thought you could shed some light on it :P
    Are there any non-integer zeros of the Riemann zeta function with real part not equal to 1/2?
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    Problem 268***

    Evaluate \displaystyle \int_0^{\infty} \frac{\sin( 2 \tan^{-1} (t) )}{(1+t^2)(e^{\pi t}+1)} \ dt

    Edit: Btw, had another look at the dreaded \frac{1}{2} (tanx cos (tanx)) integral and have made a few breakthroughs. I feel as though I might be nearing a solution but I'm not quite there yet! Has anyone else done anything substantial?
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    I am currently learning this stuff, so forgive me if I have made mistakes or chosen an odd contour.

    Solution 268

    Consider \displaystyle f(z)=\frac{2z}{(1+z^2)^2(e^{\pi z}+1)}, which has a pole of order 3 at i and simple poles at (2k+1)i for k\geq 1.

    Let \zeta be the positively oriented square formed by the lines \lambda,\omega,h,h' indicated below.


    We find that:

    \displaystyle\text{Res}(f,i)=i \left(\frac{1}{8\pi}+\frac{\pi}{  24}\right) and \displaystyle\text{Res}\left( \frac{1}{e^{\pi z}+1},(2k+1)i\right)=-\frac{1}{\pi} so \displaystyle\text{Res}(f,(2k+1)  i)=-\frac{(4k+2)i}{\pi(4k^2+4k)^2}

    Therefore, as n\to\infty:

    \begin{aligned}\displaystyle\int  _{\zeta} f(z)\,dz=2\pi i\left[i \left(\frac{1}{8\pi}+\frac{\pi}{  24}\right)-\sum_{k\geq 1}\frac{(4k+2)i}{\pi(4k^2+4k)^2} \right]=2\pi i\left[i \left(\frac{1}{8\pi}+\frac{\pi}{  24}\right)-\frac{i}{8\pi}\left]=-\frac{\pi^2}{12}

    Furthermore,

    \begin{aligned}\displaystyle \left|\int_{h}f(z)\,dz \right|\leq  \int_{0}^{1} 2n\left| f\big(n(1+2ti)\big) \right|\,dt\leq  \sqrt{5}e^{-n\pi}\to 0 and similarly \displaystyle \int_{h'} f(z)\dz\to 0 and \displaystyle \int_{w} f(z)\dz\to 0

    Thus,

    \displaystyle \int_{\lambda} f(z)\,dz\to \int_{-\infty}^{\infty} \frac{2x\,dx}{(1+x^2)^2(e^{\pi x}+1)}=-\frac{\pi^2}{12}. Last, observe that if p=\displaystyle\int_0^{\infty} f(x)\,dx and q=\displaystyle\int_{-\infty}^0 f(x)\,dx then:

    p-q=\displaystyle \int_0^{\infty} \frac{2x\,dx}{(1+x^2)^2}=1, which together with p+q=-\dfrac{\pi^2}{12} yields:

    \displaystyle\int_{0}^{\infty} \frac{2x\,dx}{(1+x^2)^2(e^{\pi x}+1)}=\frac{1}{2}-\frac{\pi^2}{24}
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    (Original post by Smaug123)
    There's a little puzzle that's been niggling at me - thought you could shed some light on it :P
    Are there any non-integer zeros of the Riemann zeta function with real part not equal to 1/2?
    :lol:

    (Original post by FireGarden)
    Exactly how much have you practiced integration? It's just silly how quickly you do basically all of them!
    Quite a bit, and thank you.
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    (Original post by Lord of the Flies)
    x
    Mind-blowing stuff as usual! How do you make your diagrams?!

    Problem 269***

    Using the previous problem as a hint,

    Express \zeta in the form \displaystyle \zeta(s)=f(s)+\int_0^{\infty} g(s,t) \ dt where f and g are combinations of 'elementary' functions and s \in \mathbb{C} \  \{1 \}.
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    (Original post by Jkn)
    Problem 268***

    Evaluate \displaystyle \int_0^{\infty} \frac{\sin( 2 \tan^{-1} (t) )}{(1+t^2)(e^{\pi t}+1)} \ dt
    Is this possible without contour integration? Contour integration is horrible
 
 
 
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