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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.
    It may be of use to note

    \displaystyle \frac{\log(1+x^n)}{1+x^n}=-\sum_{k \ge 1}H_k (-x)^{nk}

    which comes in handy in many places, for example page 7 here http://129.81.170.14/~vhm/papers_html/rmt-final.pdf . You may be able to use Ramanujan's Master theorem in those questions, although I haven't tried it and you may end up taking \Gamma(-n), n \in \mathbb{N}_0.
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    (Original post by henpen)
    It may be of use to note

    \displaystyle \frac{\log(1+x^n)}{1+x^n}=-\sum_{k \ge 1}H_k (-x)^{nk}

    which comes in handy in many places, for example page 7 here http://129.81.170.14/~vhm/papers_html/rmt-final.pdf . You may be able to use Ramanujan's Master theorem in those questions, although I haven't tried it and you may end up taking \Gamma(-n), n \in \mathbb{N}_0.
    Well now you've lost me but I look forward to seeing solution (google will have to be my friend for the strange stuff though)
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    Problem 400 **

    Prove
     \displaystyle \frac{e^{-x}+e^{-1}}{x-1}+\cosh(x)=\sum_{n \ge 0}\frac{x^ n}{n!}\left \{ \frac{n!}{e} \right \},
    where the squiggly brackets pertain to the fractional part function.
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    (Original post by Mladenov)
    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.
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    From the theory, we have \displaystyle \beta_{(1,2)}(0,1) = - \sum_{1}^{\infty} \frac{4H_{n}}{(n+1)^{3}}.


    The result in the hint is attainable using \frac{\log(1-u)}{1-u}=- \sum_{k \ge 1}H_k u^k. I've had no luck simplifying the sum, though.
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    (Original post by henpen)
    The result in the hint is attainable using \frac{\log(1-u)}{1-u}=- \sum_{k \ge 1}H_k u^k. I've had no luck simplifying the sum, though.
    Yep, I just wanted to give an example.

    To evaluate this sum, we can proceed as follows:

    Note \displaystyle H_{n} = \int_{0}^{1} \frac{1-x^{n}}{1-x}dx. Absolute convergence implies that we can interchange the order of integration and summation, and doing so we get

    \displaystyle \sum_{1}^{\infty} H_{n}\frac{1}{(n+1)^{3}} &= \int_{0}^{1} \frac{1}{1-x} \sum_{1}^{\infty} \frac{1-x^{n}}{(n+1)^{3}}dx.
    Now, integration by parts will do the job. For more general identities, which I am not keen to work out, look at this blog post.

    By the way, there is a better way to evaluate the integral; consider the function \displaystyle \int_{0}^{x} \frac{\ln (1-t)}{t}dt.
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    posted in the wrong thread sorry
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    (Original post by Mladenov)
    Problem 83**

    Let x, a, and b be positive integers such that x^{a+b}=a^{b}b. Then a=x and b=x^{x}.
    Not sure what the problem is here:

    x^{(a + b)} = x^ax^b

    x^ax^b = a^bb

    Easy to see it holds that a = x and b = x^x

    Is there anything else to do?
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    (Original post by bananarama2)
    Hint for my problem
    Consider the conservation of angular momentum.
    It lands on the tower, how far down depends on how far out you hold the ball.
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    (Original post by JamesyB)
    Not sure what the problem is here:

    x^{(a + b)} = x^ax^b

    x^ax^b = a^bb

    Easy to see it holds that a = x and b = x^x

    Is there anything else to do?
    You have to prove that there are no other possibilities, not only to check trivially that this is a solution.
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    Found it difficult to put my problem into words, if you want me to clarify something because I've explained it poorly please don't hesitate to quote/inbox me Really wanted to try posting something a bit different

    Problem 401**/***

    i^i, where i is the complex unit, is a multi-valued function that quite surprisingly always takes on real values. Each value of the function can be expressed in terms of -k, where k can be any integer.

    Find the infinite sum of the values of i^i for which k is a positive integer.
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    (Original post by Flauta)
    Found it difficult to put my problem into words, if you want me to clarify something because I've explained it poorly please don't hesitate to quote/inbox me Really wanted to try posting something a bit different

    Problem 401**/***

    i^i, where i is the complex unit, is a multi-valued function that quite surprisingly always takes on real values. Each value of the function can be expressed in terms of -k, where k can be any integer.

    Find the infinite sum of the values of i^i for which k is a positive integer.
    It's quite late, so I might well have messed something up…

    Spoiler:
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    \displaystyle i^i = \exp((\dfrac{\pi}{2}+2 n \pi) i)^i = \exp((\dfrac{\pi}{2}+2 n \pi) i \times i) = \exp(-\dfrac{\pi}{2}-2 n \pi).
    \displaystyle \sum_{n=1}^{\infty} i_n^{i_n} = \sum_{n=1}^{\infty}\exp(-\dfrac{\pi}{2}-2 n \pi) = \exp(-\dfrac{\pi}{2}) \times \sum_{n=1}^{\infty}\exp(-2 n \pi).
    But that is a geometric series, with multiplicative term \exp(-2 \pi) and initial term 1, so the infinite sum is \dfrac{1}{-1+\exp(2 \pi)} by some simple shuffling around in the infinite-sum formula.
    Hence the required sum from the question is \exp(-\dfrac{\pi}{2}) \times \dfrac{1}{-1+\exp(2 \pi)}.
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    Here are some nice and easy questions for A-level students who may find themselves lost on this thread due to us mostly considering higher level problems

    Problem 402*

    Find, for \psi, \eta \neq 0

    I = \displaystyle \int e^{\eta x} \sin(\psi x) dx

    and give a condition on x such that I>0

    Problem 403*

    Find \psi and \eta such that

    \ln(\psi + \eta) = \ln(\psi) + \ln(\eta)
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    Can't help but feel that this is a bit verbose and there should be a more direct way to evaluate this but here goes...

    Solution 394

    \displaystyle \begin{aligned} \int_0^1 \frac{\ln x \ln^2 (1-x)}{x} \ dx & \overset{x \mapsto 1-x}= \int_0^1 \frac{\ln^2 x \ln (1-x)}{1-x} \ dx \\ & =  \int_0^1 \ln (1-x) \ln^2 x \sum_{n=0}^{\infty} x^n \ dx \\ & = \sum_{n=0}^{\infty} \int_0^1 \ln (1-x) \ln^2 x \cdot x^n \ dx \\ & = -\sum_{n=0}^{\infty} \int_0^1 \ln^2 x \cdot x^n \cdot \sum_{m=1}^{\infty} \frac{x^m}{m} \ dx \\ & = - \sum_{n=0}^{\infty} \sum_{m=1}^{\infty} \int_0^1 \frac{x^m}{m} \cdot \ln^2 x \cdot x^n \ dx \\ & \overset{\text{I.B.P.}}= - 2 \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m(m+n)^3}  \end{aligned}

    Interchanging the summation bases and adding yields:

    \displaystyle\begin{aligned} -2\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m(m+n)^3} & =  - \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{mn(m+n)^2}  \\ & = -2 \sum_{k=2}^{\infty} \frac{H_{k-1}}{k^3} \\ &  = -2 \left( \sum_{k=1}^{\infty} \frac{H_k}{k^3} - \zeta (4)\right) \end{aligned}

    So it remains to evaluate \displaystyle \sum_{k=1}^{\infty} \frac{H_k}{k^3}.

    Using the relation \displaystyle H_k = \int_0^1 \frac{1-x^k}{1-x} \ dx:

    \displaystyle \begin{aligned} \sum_{k=1}^{\infty} \frac{H_k}{k^3} & = \int_0^1 \frac{1}{1-x} \sum_{k=1}^{\infty} \frac{1-x^k}{k^3} \ dx \\ & = \int_0^1 \frac{\zeta (3) - \sum_{k=1}^{\infty} \frac{x^k}{k^3}}{1-x} \ dx \\ & \overset{\text{I.B.P.}}= \underbrace{-\ln (1-x) \cdot \left(\zeta(3) - \sum_{k=1}^{\infty} \frac{x^k}{k^3}\right)\bigg|_0^1  }_{=0}  - \int_0^1 \ln (1-x) \cdot \sum_{k=1}^{\infty} \frac{x^{k-1}}{k^2} \ dx \\ & =  \int_0^1 \sum_{k=1}^{\infty} \frac{x^k}{k} \cdot \sum_{k=1}^{\infty} \frac{x^{k-1}}{k^2} \ dx \\ & = \int_0^1 \sum_{k=1}^{\infty} \frac{x^{k-1}}{k} \cdot \sum_{k=1}^{\infty} \frac{x^k}{k^2} \ dx \\ & \overset{\text{I.B.P.}}= \frac{1}{2} \left( \sum_{k=1}^{\infty} \frac{x^k}{k^2} \right)^2 \bigg|_0^1 \\ & = \tfrac{1}{2} \zeta^2 (2) \end{aligned}

    Putting everything together, we have:

    \displaystyle \int_0^1 \frac{\ln x \ln^2 (1-x)}{x} \ dx = 2 \zeta(4) - \zeta^2 (2) = \boxed{-\tfrac{1}{2} \zeta (4)}
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    (Original post by Felix Felicis)
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    Can't help but feel that this is a bit verbose and there should be a more direct way to evaluate this but here goes...

    Solution 394

    \displaystyle \begin{aligned} \int_0^1 \frac{\ln x \ln^2 (1-x)}{x} \ dx & \overset{x \mapsto 1-x}= \int_0^1 \frac{\ln^2 x \ln (1-x)}{1-x} \ dx \\ & =  \int_0^1 \ln (1-x) \ln^2 x \sum_{n=0}^{\infty} x^n \ dx \\ & = \sum_{n=0}^{\infty} \int_0^1 \ln (1-x) \ln^2 x \cdot x^n \ dx \\ & = -\sum_{n=0}^{\infty} \int_0^1 \ln^2 x \cdot x^r \cdot \sum_{m=1}^{\infty} \frac{x^m}{m} \ dx \\ & = - \sum_{n=0}^{\infty} \sum_{m=1}^{\infty} \int_0^1 \frac{x^m}{m} \cdot \ln^2 x \cdot x^n \ dx \\ & \overset{\text{I.B.P.}}= - 2 \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m(m+n)^3}  \end{aligned}

    Interchanging the summation bases and adding yields:

    \displaystyle\begin{aligned} -2\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m(m+n)^3} & =  - \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{mn(m+n)^2}  \\ & = -2 \sum_{k=2}^{\infty} \frac{H_{k-1}}{k^3} \\ &  = -2 \left( \sum_{k=1}^{\infty} \frac{H_k}{k^3} - \zeta (4)\right) \end{aligned}

    So it remains to evaluate \displaystyle \sum_{k=1}^{\infty} \frac{H_k}{k^3}.

    Using the relation \displaystyle H_k = \int_0^1 \frac{1-x^k}{1-x} \ dx:

    \displaystyle \begin{aligned} \sum_{k=1}^{\infty} \frac{H_k}{k^3} & = \int_0^1 \frac{1}{1-x} \sum_{k=1}^{\infty} \frac{1-x^k}{k^3} \ dx \\ & = \int_0^1 \frac{\zeta (3) - \sum_{k=1}^{\infty} \frac{x^k}{k^3}}{1-x} \ dx \\ & \overset{\text{I.B.P.}}= \underbrace{-\ln (1-x) \cdot \left(\zeta(3) - \sum_{k=1}^{\infty} \frac{x^k}{k^3}\right)\bigg|_0^1  }_{=0}  - \int_0^1 \ln (1-x) \cdot \sum_{k=1}^{\infty} \frac{x^{k-1}}{k^2} \ dx \\ & =  \int_0^1 \sum_{k=1}^{\infty} \frac{x^k}{k} \cdot \sum_{k=1}^{\infty} \frac{x^{k-1}}{k^2} \ dx \\ & = \int_0^1 \sum_{k=1}^{\infty} \frac{x^{k-1}}{k} \cdot \sum_{k=1}^{\infty} \frac{x^k}{k^2} \ dx \\ & \overset{\text{I.B.P.}}= \frac{1}{2} \left( \sum_{k=1}^{\infty} \frac{x^k}{k^2} \right)^2 \bigg|_0^1 \\ & = \tfrac{1}{2} \zeta^2 (2) \end{aligned}

    Putting everything together, we have:

    \displaystyle \int_0^1 \frac{\ln x \ln^2 (1-x)}{x} \ dx = 2 \zeta(4) - \zeta^2 (2) = \boxed{-\tfrac{1}{2} \zeta (4)}
    Well done.

    Problem 404***

    Evaluate \displaystyle \int_{0}^{\infty} \frac{\sqrt{x} \ln x}{1+x^{2}}dx.
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    (Original post by Indeterminate)
    Problem 402*

    Find, for \psi, \eta \neq 0

    I = \displaystyle \int e^{\eta x} \sin(\psi x) dx

    and give a condition solely dependent on x such that I>0
    Solution-ish 402* :ninja:

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    IBP 1
    \begin{array}{ccc} u = \sin(\psi x) & \implies & \dfrac{du}{dx} = \psi \cos(\psi x) \\ \dfrac{dv}{dx} = e^{\eta x} & \implies & v = \dfrac{e^{\eta x}}{\eta} \end{array}


    \displaystyle \begin{aligned} I & = \int e^{\eta x} \sin(\psi x)\ dx \\ & \overset{\text{I.B.P}}= \ \frac{1}{\eta} \  e^{\eta x} \sin(\psi x) - \frac{\psi}{\eta} \int e^{\eta x} \cos(\psi x)\ dx \end{aligned}

    IBP 2
    \begin{array}{ccc} u = \cos(\psi x) & \implies & \dfrac{du}{dx} = -\psi \sin(\psi x) \\ \dfrac{dv}{dx} = e^{\eta x} & \implies & v = \dfrac{e^{\eta x}}{\eta} \end{array}


    \displaystyle \begin{aligned} I & \ \overset{\text{I.B.P}}= \ \frac{1}{\eta} \ e^{\eta x} \sin(\psi x) - \frac{\psi}{\eta^2} \ e^{\eta x} \cos(\psi x) - \frac{\psi^2}{\eta^2} \ I \\ & \  \ = \ \dfrac{e^{\eta x} \Big( \eta \sin(\psi x) - \psi \cos(\psi x) \Big)}{\psi^2 + \eta^2}

    Now for finding the condition for which the integral is greater than 0:

    \eta \sin(\psi x) - \psi \cos(\psi x) \ \equiv \ \sqrt{\psi^2 + \eta^2} \sin \left( \psi x - \arctan \left( \frac{\psi}{\eta} \right) \right)

    \therefore \ I = \underbrace{\dfrac{e^{\eta x}}{\sqrt{\psi^2 + \eta^2}}}_{{\forall x: \ e^{\eta x} > 0}} \ \sin \left( \psi x - \arctan \left( \frac{\psi}{\eta} \right) \right)

    Since the term involving e^{\eta x} is positive for all real x, we must only consider the values of x for which the sine term is greater than 0. Our problem is now reduced to:

    \sin \left( \psi x - \arctan \left( \frac{\psi}{\eta} \right) \right) > 0

    I decided to take the graphical approach
    The graph of \sin x is first translated horizontally by \arctan \left( \frac{\psi}{\eta} \right) units.

    This graph is then horizontally compressed by a scale factor of \psi.

    It's period is also altered and is now \frac{2n\pi}{\psi}.

    Accordingly, we can now see that the values of x for which the sine term above is greater than 0 are:

    \dfrac{ \arctan \left( \frac{\psi}{\eta} \right) + 2n\pi}{\psi} < x < \dfrac{ \arctan \left( \frac{\psi}{\eta} \right) + \pi + 2n\pi}{\psi}
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    (Original post by Mladenov)
    Well done.

    Problem 404***

    Evaluate \displaystyle \int_{0}^{\infty} \frac{\sqrt{x} \ln x}{1+x^{2}}dx.
    Substitute \sqrt{x}=u, and using that the integrand is now even,

    \displaystyle I= \int_{-\infty}^{\infty} \frac{ \ln (u^2)}{1+u^{4}}du.

    The integrand has simple poles at the fourth roots of unity, [\zeta^{k}_4]_{1\le k \le 4}. Make a semicircular contour, integrating anticlockwise, and noting that the curvy bit tends to 0 as the radius of the semicircle \rightarrow \infty. Thus

    \displaystyle \int_{-\infty}^{\infty} \frac{ \ln (u^2)}{1+u^{4}}du=2\pi i (Res[\frac{\ln(z^2)}{1+z^4}, z=\zeta^1_4]+Res[\frac{\ln(z^2)}{1+z^4}, z=\zeta^2_4])

    \displaystyle =2\pi i \left( \frac{2\ln(\zeta^1_4)}{(\zeta^1_  4-\zeta^2_4)(\zeta^1_4-\zeta^3_4)(\zeta^1_4-\zeta^4_4)}+ \frac{2\ln(\zeta^2_4)}{(\zeta^2_  4-\zeta^1_4)(\zeta^2_4-\zeta^3_4)(\zeta^2_4-\zeta^4_4)} \right)

    \displaystyle =2\pi i \left( \frac{2\frac{i\pi}{4}}{(\sqrt{2}  )(\sqrt{2}(1+i))(\sqrt{2}i)}+ \frac{2\frac{i 3\pi}{4}}{(-\sqrt{2})(\sqrt{2}i)(-\sqrt{2}(-1+i))} \right)

    \displaystyle =2\pi i \left( \frac{\pi(1-i)}{8 \sqrt{2}}+\frac{\pi(-1-i)}{8 \sqrt{2}}\right)=\frac{\pi^2}{2 \sqrt{2}}.

    What's incorrect with this method?
    Solution 2: A real method.

    \displaystyle I=4J=2 \int_0^\infty \frac{2 \ln(u)}{1+u^4}du.

    \displaystyle J = \int_0^1 \frac{ \ln(u)}{1+u^4}du+ \int_1^\infty  \frac{ \ln(u)}{1+u^4}du

    \displaystyle = \int_0^1 \ln(u) \sum_{k \ge 0} (-1)^ku^{4k}du+\int_1^\infty \ln(u) \sum_{k \ge 0}(-1)^ku^{-4(k+1)}du

    \displaystyle =- \sum_{k \ge 0} \frac{(-1)^k}{(4k+1)^2}+\sum_{k \ge0} \frac{(-1)^k}{(4k+3)^2}=G

    Where G is Catalan's constant. This answer is incorrect, but I cannot see what's incorrect in the method.
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    Problem 405**

    Evaluate

    \displaystyle \int_0^\pi \frac{x \cos(x)}{1+\sin^2(x)}dx.

    Problem 406**

    Find a closed form (i.e. no series) for

    \displaystyle \sum_{k \ge 1} \frac{\zeta(2k+1)-1}{k+1}.
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    (Original post by henpen)
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    Substitute \sqrt{x}=u, and using that the integrand is now even,

    \displaystyle I= \int_{-\infty}^{\infty} \frac{ \ln (u^2)}{1+u^{4}}du.

    The integrand has simple poles at the fourth roots of unity, [\zeta^{k}_4]_{1\le k \le 4}. Make a semicircular contour, integrating anticlockwise, and noting that the curvy bit tends to 0 as the radius of the semicircle \rightarrow \infty. Thus

    \displaystyle \int_{-\infty}^{\infty} \frac{ \ln (u^2)}{1+u^{4}}du=2\pi i (Res[\frac{\ln(z^2)}{1+z^4}, z=\zeta^1_4]+Res[\frac{\ln(z^2)}{1+z^4}, z=\zeta^2_4])

    \displaystyle =2\pi i \left( \frac{2\ln(\zeta^1_4)}{(\zeta^1_  4-\zeta^2_4)(\zeta^1_4-\zeta^3_4)(\zeta^1_4-\zeta^4_4)}+ \frac{2\ln(\zeta^2_4)}{(\zeta^2_  4-\zeta^1_4)(\zeta^2_4-\zeta^3_4)(\zeta^2_4-\zeta^4_4)} \right)

    \displaystyle =2\pi i \left( \frac{2\frac{i\pi}{4}}{(\sqrt{2}  )(\sqrt{2}(1+i))(\sqrt{2}i)}+ \frac{2\frac{i 3\pi}{4}}{(-\sqrt{2})(\sqrt{2}i)(-\sqrt{2}(-1+i))} \right)

    \displaystyle =2\pi i \left( \frac{\pi(1-i)}{8 \sqrt{2}}+\frac{\pi(-1-i)}{8 \sqrt{2}}\right)=\frac{\pi^2}{2 \sqrt{2}}.
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    Alternate solution to 404 (still can't do contour integration :nopity:)

    We wish to evaluate:

    \displaystyle\int_0^{\infty} \frac{\sqrt{x} \ln x}{1+x^2} \ dx

    Start by letting x \mapsto \tan x :

    \displaystyle\begin{aligned} \int_0^{\infty} \frac{\sqrt{x} \ln x}{1+x^2} \ dx & = \int_0^{\frac{\pi}{2}} \sqrt{\tan x} \ln \tan x \ dx \\ & = \underbrace{\int_0^{\frac{\pi}{2  }} \sqrt{\tan x} \ln \sin x \ dx}_{I} - \underbrace{\int_0^{\frac{\pi}{2  }} \sqrt{\tan x} \ln \cos x \ dx}_{J} \end{aligned}

    \bullet \quad I: For \alpha > 0, define

    \displaystyle\begin{aligned}K := \text{B}\left(1 - \tfrac{\alpha}{4}, \tfrac{1}{4}\right) & = \int_0^1 x^{-\frac{\alpha}{4}} (1-x)^{-\frac{3}{4}} \ dx \\ & \overset{x \mapsto \sin^2 x}= 2 \int_0^{\frac{\pi}{2}} \sin^{\frac{-\alpha + 1}{2}} x \sqrt{\tan x} \ dx \\ \implies \frac{\partial K}{\partial \alpha} \bigg|_{\alpha = 1} & = - I\end{aligned}

    \displaystyle\frac{\partial K}{\partial \alpha} \bigg|_{\alpha = 1} = \frac{\sqrt{2}\pi}{4} \left(3 \ln 2 - \frac{\pi}{2}\right) \implies I = \frac{\sqrt{2}\pi}{4} \left(\frac{\pi}{2} - 3 \ln 2\right)

    \bullet \quad J: Now, for \beta > 0, define

    \displaystyle\begin{aligned} L := \text{B} \left(\tfrac{3}{4}, 1 - \tfrac{3\beta}{4}\right)  & = \int_0^1 x^{-\frac{1}{4}} (1-x)^{-\frac{3\beta}{4}} \ dx \\ & \overset{x \mapsto \sin^2 x}= 2 \int_0^{\frac{\pi}{2}} \sqrt{\tan x} \cos^{\frac{3}{2} (1 - \beta)} x \ dx \\ \implies \frac{\partial L}{\partial \beta}\bigg|_{\beta = 1} & = -3J\end{aligned}

    \displaystyle \begin{aligned} \frac{\partial L}{\partial \beta} \bigg|_{\beta = 1} = \frac{3\sqrt{2}\pi}{4}\left( \frac{\pi}{2} + 3 \ln 2\right) \implies J = - \frac{\sqrt{2}\pi}{4} \left( \frac{\pi}{2} + 3 \ln 2\right)\end{aligned}

    Bringing everything together:

    \displaystyle \int_0^{\infty} \frac{\sqrt{x} \ln x}{1+x^2} \ dx = I - J = \frac{\sqrt{2}\pi^2}{4}
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    Looks like my system of differential equations was a bit beyond you guys I guess. I'll post the solution in the very near future.

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