3

Proposed:

$$\int_{-\alpha}^{\alpha}{x^k}\sqrt{\alpha+x\over \alpha-x}\ln^n\left({\alpha+x\over \alpha-x}\right)\mathrm dx=F(k,n)\tag1$$

Let $k=-1$ and $F(-1,n)=F(n)$

$$\int_{-\alpha}^{\alpha}{1\over x}\sqrt{\alpha+x\over \alpha-x}\ln^n\left({\alpha+x\over \alpha-x}\right)\mathrm dx=(-\pi)^{n+1}F(n)\tag2$$ Where $n,\alpha\ge1$

We have the following

$n=1\implies$ $F(1)=1$

$n=2\implies$ $F(2)=-1$

$n=3\implies$ $F(3)=2$

$n=4\implies$ $F(4)=-5$

$n=5\implies$ $F(5)=16$

$n=6\implies$ $F(6)=-61$

How can we find the closed form of $(2)?$

$u={\alpha+x\over \alpha-x}\implies dx={(\alpha-x)^2\over 2\alpha}$

$x=\alpha\cdot{u-1\over u+1}$

$(\alpha-x)^2={4\alpha^2\over (u+1)^2}$

$(2)\implies$

$$2\int_{0}^{\infty}{\sqrt{u}\over u^2-1}\cdot\ln^n(u)\mathrm du\tag3$$

$$....$$

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    Very much reminds me of this. – Simply Beautiful Art Jun 12 '17 at 11:28
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    First of all, the $\alpha$ is irrelevant:

    $$\int_{-\alpha}^{+\alpha}{x^k}\sqrt{\alpha+x\over \alpha-x}\ln^n\left({\alpha+x\over \alpha-x}\right)~\mathrm dx=\alpha^{k+1}\int_{-1}^{+1}x^k\sqrt{\frac{1+x}{1-x}}\ln^n\left(\frac{1+x}{1-x}\right)~\mathrm dx$$

    Secondly, consider this integral:

    $$I_k(t)=\int_{-1}^{+1}x^k\left(\frac{1+x}{1-x}\right)^t~\mathrm dx$$

    and note that

    $$\frac{\partial I_k(t)}{\partial t^n}\bigg|{t=1/2}=\int{-1}^{+1}x^k\sqrt{\frac{1+x}{1-x}}\ln^n\left(\frac{1+x}{1-x}\right)~\mathrm dx$$

    Those are my thoughts :-)

     – Simply Beautiful Art Jun 12 '17 at 11:39
  • I dont know if it helps but a good start would be $x=a\cos (u) $ – Archis Welankar Jun 12 '17 at 11:48
  • Oops, it should be $\frac{\partial^nI_k(t)}{\partial t^n}$ at the end there. – Simply Beautiful Art Jun 12 '17 at 11:55
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    It is full of integrals with irrelevant parameters here on MSE. If you want to ask about $\int_{0}^{1}\frac{u^{1/2}\log^n(u)}{1-u^2},du$, just ask about it, but please add your attempts, since this integral can be tackled by the same techniques solving a good $95%$ of your questions. – Jack D'Aurizio Jun 12 '17 at 16:48
  • Sorry everyone I ask a wrong question. I was looking for the closed form of (2) not (1) –  Jun 12 '17 at 17:47
  • Mr jack I have been here on MSE for almost 8 months, wasting a lot of time. Anyway see you in the future when I get my degree in maths. –  Jun 12 '17 at 18:20
  • @Jackiechan, I don't consider this a waste of time. – Zaid Alyafeai Jun 13 '17 at 02:45
  • Thank you @Zaid Alyafeai, soon it comes to an end here for me. Wish you all the best in the future. Thank once again. –  Jun 13 '17 at 07:48
  • @Jackiechan, you just need to do what feels good for you. If you enjoy what you are doing then continue it regardless of what people think. Wish you the best. – Zaid Alyafeai Jun 13 '17 at 17:31

1 Answers1

1

The comment of Simply Beautiful Art gives a big hint on how to proceed. I cannot give a general formula to (1), although it is possible albeit complicated.

Let $$I(k,t)=\int_{-1}^{+1}x^k\left(\frac{1+x}{1-x}\right)^t dx \quad\quad J(k,n)=\int_{-1}^{+1}x^k\sqrt{\frac{1+x}{1-x}}\ln^n\left(\frac{1+x}{1-x}\right)dx$$ making $u = \frac{1+x}{1-x}$ gives $$I(k,t) = 2\int_{0}^{\infty} \frac{u^t(u-1)^k}{(u+1)^{2+k}} dx$$

When $k$ is a positive integer, we can expand and note that $$\int_{0}^{\infty} \frac{u^a}{(u+1)^b} du = B(a+1,b-a-1)$$ with the beta function.

Since the derivative of gamma function at half-integer value can be calculated, the closed form for each $J(k,n)$ can also be found.

For instances

$$J(1,1) = 2\pi \quad J(2,1) = \frac{5\pi}{3} \quad J(2,1) = \frac{5\pi}{3} \quad \\ J(4,1) = \frac{89\pi}{60} \quad J(5,1) = \frac{89\pi}{60} \quad J(6,1) = \frac{381\pi}{280}$$

$$\begin{aligned} J(1,2) = J(2,2) = 4\pi + \frac{\pi^3}{2} \\ J(3,2) = J(4,2) = \frac{14\pi}{3} + \frac{3\pi^3}{8} \\ J(5,2) = J(6,2) = \frac{439\pi}{90} + \frac{5\pi^3}{16} \\ \end{aligned}$$

$$\begin{aligned} J(1,3) & = 6\pi^3 \\ J(2,3) &= J(3,3) = 8\pi + 5\pi^3 \\ J(4,3) &= J(5,3) = 12\pi + \frac{89\pi^3}{20} \\ \end{aligned}$$

pisco
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