81

What tools or ways would you propose for getting the closed form of this integral?

$$\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$$

EDIT: It took a while since I made this post. I'll give a little bounty for the solver of the problem, 500 points bounty.

Supplementary question:

Calculate

$$\int_0^{\pi/4}\frac{\log(1-x)\log(x)\log(1+x) \tan^2(x)}{1-x\tan^2(x)} \ dx$$

mathreadler
  • 25,824
user 1591719
  • 44,216
  • 12
  • 105
  • 255
  • 6
    Could you tell what you tried in order we avoid wrong tracks ? Thanks and cheers :) – Claude Leibovici Jul 16 '14 at 16:34
  • @ClaudeLeibovici Sure. I thought of using the differentiation under the integral sign, but no success so far. – user 1591719 Jul 16 '14 at 16:37
  • It is possible to solve it by series; since $ |x\tan^2x|<1$ then expand $\frac {\tan^2x}{1-x\tan^2x }$ by geometric series, similarly write the series of $\log. $ then multiplying the series. . . – Mohammad W. Alomari Jul 19 '14 at 13:18
  • Have you tried formulating it as a complex integral and using the Residue theorem? – artificial_moonlet Aug 19 '14 at 15:29
  • @ artificial_moonlet, it has no poles. – Amad27 Jan 01 '15 at 08:24
  • I started with visualizing it first using Wolfram Alpha and saw that ugliness that might make it tricky... – wigy Jan 26 '15 at 09:27
  • 11
    I highly doubt this has a closed form. For one, the choice of $\pi/4$ as an upper limit seems arbitrary, since the integrand has no pole at $\pi/4$. As well, integrals that involve $x\tan{x}$ or $\ln(x)\tan{x}$ are hard enough separately (for example even something like $\int_0^{\pi/4}x^n\tan{x}\text{d}x$ only has a closed form in terms of an infinite sum of zeta functions). I've messed around with this for about a day, and it seems you either have to eliminate the $\ln(1-x)\tan^2x$ and the $x\tan^2x$ simultaneously (which I've to be found very hard), or deal with messy nested infinite sums. – user3002473 Jul 28 '15 at 17:11
  • Does $\log$ denote the natural logarithm or common logarithm? I've seen it used both ways. – Franklin Pezzuti Dyer Apr 27 '17 at 19:57
  • @FranklinP.Dyer natural logarithm – user 1591719 Apr 27 '17 at 20:25
  • Taylor expansion of 1 - x is 1, which is appropriate for tangent function. The log of 1 - x is the log of x for x < 1. Which gives you logx which is 1/x summed between 0 and pi over four gives 4 over pi. – McTaffy Jul 25 '17 at 15:17
  • 17
    What is so interesting about this integrals ? – user90369 Dec 05 '17 at 12:52
  • 5
    The fact that no one has cracked it in the last three plus years means its very hard to do, and consequently, intrinsically interesting. – omegadot Dec 07 '17 at 01:04
  • 5
    @omegadot : No, it just means that the probability is very high that there is no closed form. (You can immediately construct endless many integrals without any closed form.) But where does such a problem come ? This could be interesting. – user90369 Dec 07 '17 at 10:04
  • 2
    We need the help of Cleo https://math.stackexchange.com/users/97378/cleo – Jürgen Sukumaran Feb 19 '18 at 22:53
  • Considering how readily Wolfram pumps out an approximate I'd hedge a bet that there is a closed form, at least in some sense of the word. – George C Feb 20 '18 at 10:38
  • 1
    I use the Mathematica to solve this integral, the result is -0.463066. – Wei-Cheng Liu Feb 24 '18 at 03:52
  • 1
    Try to give a closed form of $\int e^{-x^2} dx$ – Fibonacci Feb 28 '18 at 15:54
  • This problem can best be solved using a combination of the bounds trick property f(a+b-x) = f(x) and Differentiation Under The Integral. You can successfully get rid of the numerator and convert tan^2 into terms of sin(2x) but the denominator remains so ugly that I find it impossible to evaluate this integral. –  Mar 13 '18 at 06:06
  • Has anyone tried to compute it to 5000 digits and use the PSQL algorithm? – user14717 Apr 26 '18 at 01:36
  • 1
    @Waiting ,Are you sure that there is no typo and denominator is not $(1-x) tan^2 (x) $? – sirous May 10 '18 at 20:07
  • @sirous I'm sure all is perfect. – user 1591719 May 10 '18 at 20:11
  • 1
    As I've already told you and many others before, this type of definite integrals can only generate meaningful closed form expressions when evaluated over their “natural” domain of definition. However, $~\displaystyle\int_0^\alpha\frac{\ln(1-x)}{\cot^2x-x}~dx~$ diverges over $(0,~\alpha),~$ where $\alpha\simeq0.83147217989111430\ldots$ is the solution to the transcendental equation $x=\cot^2x.~$ – Lucian May 12 '18 at 19:28
  • The same observation also applies to $\displaystyle\int_0^1\frac{\ln(1-x)}{\cot^2\Big(\tfrac\pi4~x\Big)-x}~dx~$ and $~\displaystyle\int_0^\tfrac\pi4\frac{\ln\Big(1-\tfrac4\pi~x\Big)}{\cot^2x-\tfrac4\pi~x}~dx.~$ – Lucian May 12 '18 at 19:37
  • @Lucian I didn't add in my question the word meaningful, no matter what that would ever mean. You might like to look, for example, over $\int_0^{\pi/3} \log(1-x) \sin(x) \textrm{d}x$, which can definitely posses a closed-form easy to obtain. Now, in terms of evaluation integrals and series, on MSE ( and to quote you now), you and many others only saw an extremely tiny bit in terms of art of calculations. The rabbit hole is like a black hole. – user 1591719 May 12 '18 at 19:45
  • @Waiting: Exceptions confirm the rule $($of thumb$)$. – Lucian May 12 '18 at 19:48
  • @Waiting: My only meaningful suggestion would be to cross-post these questions on Math Overflow. – Lucian May 12 '18 at 19:58
  • @Lucian Regarding such calculations, I don't think there is much difference between MSE and MathOverflow. – user 1591719 May 13 '18 at 20:01
  • @Waiting: I wouldn't be so sure about that... – Lucian May 14 '18 at 01:43
  • @Lucian this is a piece of cake compared with some serious integrals and series. Forget about it. – user 1591719 May 14 '18 at 06:52
  • @Waiting: All I'm saying is, something which went unsolved at Math Stack Exchange for a semester found a solution at Math Overflow within a month. – Lucian May 14 '18 at 07:57
  • @Waiting: If you could by any chance provide the motivation behind this integral, it might help reopening its cross-post over at Math Overflow. – Lucian May 27 '18 at 13:27
  • @Lucian your problem is that you talk too much and work too less. ;) Forget about MathOverflow, they will find, invent any reason to close the question within seconds. – user 1591719 May 29 '18 at 10:22
  • @Waiting: $(1).$ The reactions I received from the Math Overflow community are basically the same as my own. In a certain sense, they did answer my question. $(2).$ I personally do not possess the mental capacity to solve this integral, even if I were to already know its closed form expression, which I doubt it has. $(3).$ All the integrals I'm interested in make sense. This one doesn't, for reasons already explained. – Lucian May 30 '18 at 03:30
  • @Lucian get used to the fact that mathematics is sometimes simply mind-blowing. You'll continue to see such integrals and series. ;) The fact that MSE and MathOverflow don't answer them doesn't mean to me that they do not make sense, or that they aren't solvable. – user 1591719 May 30 '18 at 11:56
  • @Waiting: Where did you get this integral ? Do you know of a closed form for it ? What is the meaning of the function in the integrand and/or denominator ? – Lucian May 30 '18 at 12:39
  • @user23571113, have you tried symbolic integration with Mathematica? In case no result is given, it's very likely no close form actually exists – EmG Jul 10 '18 at 13:23
  • 1
    @EmG the fact that Mathematica cannot give a closed form does not mean that it doesn't exist. There are many "difficult" integrals that actually has a closed form to which Mathematica cannot answer. Though regardless, it is still very likely there is no closed form. – John Glenn Jul 15 '18 at 10:58
  • @JohnGlenn I totally agree with you – EmG Jul 16 '18 at 10:58
  • 2
    I can't believe this question has that many upvotes also it is not confirmed that even the closed for exits – Deepesh Meena Sep 05 '18 at 14:14
  • Trying Wolfram Alpha a free account will not give the computation time (if this is possible to calculate it) but if anyone has a pro account they could try that. https://www.wolframalpha.com/input/?i=integrate+(log(1-x)tan%5E2(x))%2F(1-xtan%5E2(x))+from+0+to+pi%2F4 .Also the numerical approximation does not seem to be a recognised number, so does this not suggest that the integral has no closed form? – Henry Lee Sep 17 '18 at 13:37
  • @ForeverInactive , if you allow me I can possibly give a closed form using a little approximation. – Awe Kumar Jha Oct 28 '18 at 10:04
  • At $x=π/4$ $\frac {\log (1-x)}{cot^2(x) - x}$ becomes $\frac {\log (1-x)}{1-x}$ which gives some hope. – Awe Kumar Jha Oct 29 '18 at 03:39

1 Answers1

5

Just a few notes for a series development, because a "closed" formula is very unlikely.

$$\int\limits_0^{\pi/4} \frac{\ln(1-x) \tan^2 x}{1-x\tan^2 x} dx = -\sum\limits_{n=0}^\infty \sum\limits_{k=0}^n \frac{1}{n-k+1} \int\limits_0^{\pi/4} x^{n+1} (\tan x)^{2k+2} dx$$

Or using $\int\limits_0^{\pi/4} \frac{\ln(1-x)}{x (1-x\tan^2 x)} dx$ it becomes a little more handsomely:

$$ \int\limits_0^{\pi/4} \frac{\ln(1-x) \tan^2 x}{1-x\tan^2 x} dx = \text{Li}_2\left(\frac{\pi}{4}\right) -\sum\limits_{n=0}^\infty \sum\limits_{k=0}^n \frac{1}{n-k+1} \int\limits_0^{\pi/4} x^n (\tan x)^{2k} dx$$

Then we have: $$ \int\limits_0^{\pi/4} x^n (\tan x)^{2k} dx = (-1)^k \left(\frac{\pi^{n+1}}{(n+1)4^{n+1}} + A_{n,k} - B_{n,k} - C_{n,k} \right) \enspace$$ for:

$$A_{n,k} := \left(\sin \frac{\pi n}{2}\right)\frac{n!}{2^{n+1}} \sum\limits_{~j=0 \\ j~\text{odd}}^{2k-1} c_{2k-1,j}~\eta(n-j+1)\\ B_{n,k} := \frac{1}{2^{2n+2}}\sum\limits_{~v=0 \\ v~\text{odd}}^n \left(\sin \frac{\pi v}{2}\right)\frac{n!\pi^{n-v}}{(n-v)!} \sum\limits_{~j=0 \\ j~\text{odd}}^{2k-1} 2^j c_{2k-1,j}~\eta(v-j+1)\\ C_{n,k} := \frac{1}{2^{2n+1}}\sum\limits_{~v=0 \\ v~\text{even}}^n \left(\cos \frac{\pi v}{2}\right)\frac{n!\pi^{n-v}2^v}{(n-v)!} \sum\limits_{~j=0 \\ j~\text{odd}}^{2k-1} c_{2k-1,j}~\beta(v-j+1)\\ c_{n,j} := \frac{1}{n!} \sum\limits_{v=0}^{n+1} {\binom {n+1} v} \sum\limits_{l=j}^n \begin{bmatrix}{n+1}\\{l+1}\end{bmatrix}{\binom l j}(-v)^{l-j}$$

and the Stirling numbers of the first kind $\begin{bmatrix}n\\k\end{bmatrix}$ defined by: $$\sum\limits_{k=0}^n\begin{bmatrix}n\\k\end{bmatrix}x^k:=\prod\limits_{k=0}^{n-1}(x+k),$$

so that $ c_{ \text{odd},\text{even} }=0 $ and $ c_{ \text{even},\text{odd} }=0$. Some values $\,c_{n,j}\,$ can be seen here .

The needed analytical continuation for Dirichlet eta function $\eta(s)$ and Dirichlet beta function $\beta(s)$ can be seen in my answer of the question here . With the additional information

$$ \sum\limits_{n=0}^\infty\frac{\pi^{n+1}}{(n+1) 4^{n+1}}\sum\limits_{k=0}^n \frac{(-1)^k}{n-k+1} = \\ \text{Li}_2\left(\frac{\pi}{4}\right) + \text{Li}_2 \left(\frac{1}{2} -\frac{\pi}{8} \right) + \left(\ln\left(\frac{1}{2} +\frac{\pi}{8} \right)\right)\ln\left(1-\frac{\pi}{4}\right)$$

we get:

$$ \int\limits_0^{\pi/4} \frac{\ln(1-x) \tan^2 x}{1-x\tan^2 x} dx = \\ -\text{Li}_2 \left(\frac{1}{2} -\frac{\pi}{8} \right) - \left(\ln\left(\frac{1}{2} +\frac{\pi}{8} \right)\right)\ln\left(1-\frac{\pi}{4}\right) - \sum\limits_{n=0}^\infty \sum\limits_{k=0}^n \frac{(-1)^k}{n-k+1}(A_{n,k} - B_{n,k} - C_{n,k} )$$

user90369
  • 11,518