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Given that the closed form exist, evaluate the following Integral:

$$\int\limits_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx $$

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Aman Rajput
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    "Given"? By whom? Besides, how knowing the (indefinite, I supose) integral has a closed form can possibly help us here? Can you give more context, where does this come from, what have you achieved so far? – DonAntonio Jun 13 '16 at 17:26
  • What kind of context is missing?? @joan – Aman Rajput Jun 14 '16 at 08:10
  • @AmanRajput One that will make clear why knowing that a closed form for the (indefinite, I presume) integral exists allows us to evaluate the given definite integral. As you can see in the only answer to your question, there doesn't seem to exist an argument of the form "...and since there exists a closed form then we can deduce that...". It uses some other heavy machinery to tackle that integral. – DonAntonio Jun 14 '16 at 08:15
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    @Joanpemo: anyway, if a sledgehammer solves the problem, it does not mean a screwdriver cannot, it would be interesting to see if other, more elementary solutions exist. – Jack D'Aurizio Jun 14 '16 at 11:58
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    @AmanRajput, the context missing is described exactly in the yellow box above. You can also visit the help center. You just posted an integral with no additional information. This site is not a context ground, as was said a lot of times, it is not intended to post challenge problems. If you personally need (or want) help with this integral, you need to expand on the background – Yuriy S Jun 14 '16 at 12:27
  • @JackD'Aurizio Thank you, I know that. It is only the phrasing of the first line in the question that bewilders me: how "knowing a closed form exists" will help me to solve that ? – DonAntonio Jun 14 '16 at 13:44
  • @Joanpemo: I agree that the intro is quite irrelevant, but it is a pity that the OP didn't show his efforts, since it is an interesting integral. Ahmed's integral and its generalizations have been a successful topic here on MSE. I hope the OP will salvage his question. – Jack D'Aurizio Jun 14 '16 at 15:22
  • @JackD'Aurizio For what it's worth, this integral appeared in my attempt of solving a summation question on brilliant.org. Here's my attempt. Both forms match numerically. – MathGod Jun 15 '16 at 05:28
  • @JackD'Aurizio Looking at the form in OPs summation question here, the integral can be expressed in terms of logarithm of some irrational number and catalan's constant. – MathGod Jun 15 '16 at 05:32
  • @JackD'Aurizio Is it known that imaginary part of dilogarithms can be expressed in simpler terms or is it a new identity (assuming OP's closed form for the summation question is correct). One more thing I didn't mention, this integral and the integral in my attempt are the same since it can be obtained through IBP. – MathGod Jun 15 '16 at 05:42

2 Answers2

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Since: $$ \frac{z+4}{z^2-12 z+16} = \frac{\bar{\varphi}}{z-(2\bar\varphi)^2}+\frac{\varphi}{z-(2\varphi)^2}\tag{1}$$ the integral can be computed by exploiting the relations between the dilogarithm and the golden ratio, namely Landen's identity. A first step of partial fraction decomposition followed by integration by parts leads to:

$$ I = \frac{\pi\log 5}{8}+\frac{1}{4}\int_{0}^{\pi/2}\log\left(\frac{4-2\sin(t)-\sin^2(t)}{4+2\sin(t)-\sin^2(t)}\right)\,dt \tag{2}$$ and if we replace $t$ with $\frac{\pi}{2}-2\arctan u$ we get: $$ I = \frac{\pi\log 5}{8}+\frac{1}{2}\int_{0}^{1}\log\left(\frac{1+10v^2+5v^4}{5+10v^2+v^4}\right)\frac{dv}{1+v^2}\tag{3}$$ where the last integral is a dilogarithmic integral, namely the imaginary part of $$\text{Li}_2\left[i\left(1-\sqrt{5}+\sqrt{5-2 \sqrt{5}}\right)\right]+\text{Li}_2\left[i\left(1+\sqrt{5}-\sqrt{5+2 \sqrt{5}}\right)\right]\tag{4}$$ that is a special value of the Clausen function $\text{Cl}_2$. It is also useful to notice that: $$ \frac{1+10v^2+5v^4}{5+10v^2+v^4}=\frac{v}{\tanh(5\text{arctanh}(v))}. \tag{5}$$

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Jack D'Aurizio
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    $(5)$ surprised me the most. Is there some way to quickly 'see' such identities? – Yuriy S Jun 13 '16 at 18:56
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    @YuriyS: through an extensive use of Chebyshev polynomials :D – Jack D'Aurizio Jun 13 '16 at 18:58
  • I'm also struck by that last identity, mainly because the LHS is of the form $\dfrac{v^4 P_4(1/v)}{P_4(v)}$ where $P_4(v)$ is an even quartic polynomial. Presumably there's a good reason why that form occurs (it implies a certain kind of symmetry under $v\to 1/v$) but it's not entirely obvious. – Semiclassical Jun 13 '16 at 20:16
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    @YuriyS: $$\tanh(a\tanh^{-1}x)=\frac{(1+x)^a-(1-x)^a}{(1+x)^a+(1-x)^a}=\frac{\binom a1x+\binom a3x^3+\binom a5x^5+\cdots}{\binom a0+\binom a2x^2+\binom a4x^4+\cdots}.$$ –  Jun 13 '16 at 20:18
  • @JackD'Aurizio Can you please elaborate how does Chebyshev Polynomials come into play here? – MathGod Jun 14 '16 at 07:53
  • @IshanSingh: compute $ix U_4(ix)$ and $T_5(ix)$ or, equivalently, see Rahul's comment above. – Jack D'Aurizio Jun 14 '16 at 11:55
  • @JackD'Aurizio Thanks! – MathGod Jun 15 '16 at 14:15
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Let $J(a)=\int_0^\infty\frac1{1+t^2} \ln \frac{t^2+2t\sin 2a+1}{t^2-2t\sin 2a+1}dt $ and establish $J’(a)=8a\csc(2a)$. Note\begin{align} &I=\int_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx =\frac14\int_{0}^{1}\sin^{-1}x \ d\left(\ln \frac{4+2x-x^2}{4-2x-x^2} \right) \end{align} Integrate by parts and then substitute $x=\frac{2t}{1+t^2}$ to obtain \begin{align} I= &\ \frac\pi8\ln5-\frac14(J(\frac{3\pi}{20})-J(\frac{\pi}{20})) =\frac\pi8\ln5 -\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}2a\csc(2a)da\\ \overset{ibp} =&\ \frac\pi8\ln5 - a \ln\tan a\bigg|_{\frac{\pi}{20}}^{\frac{3\pi}{20}} +\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}\ln \tan a\ da\\ =&\ \frac\pi8\ln5 - \frac{3\pi}{20}\ln \tan \frac{3\pi}{20} +\frac{\pi}{20}\ln \tan \frac{\pi}{20}-\frac25 G\\ \end{align} where$\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}\ln \tan a\ da =-\frac25G$

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