5
$\begingroup$

As I do not know the complex behavior of this function, it would be even harder to integrate past the real domain. The upper bound for the domain is a constant I will denote β.

$${{Q_2}=\int_0^βQ^{-1}(x,x,x)dx= \int_0^β Q^{-1}\left(x,\frac{Γ(x,x)}{Γ(x)}-x\right)dx= \int_0^β Q^{-1}\left(x,Q(x,x)-x\right)dx=.221221...}$$

Here is the graph of the constant:

enter image description here Here is one of my sources of this function here. I very strongly suggest it to learn more about this function.

Here is also the proposed defined upper domain bound to this function. This is the graph of the reciprocal integrand. Let this aforementioned point be (β,0). Notice how the graph is 0 for x=βenter image description here

Related

Here is the value I have for beta using this method:β=$.259913533314486931…\ $.

The upper bound of the integral is the fixed point for $Q^{-1}(x,x)$ and Q(x,x). The constant also seems to be defined as the maximum value of the same function.

This means that: $${Q^{-1}(β,β,β)= Q^{-1}(β,Q(β,β)-β)=Q^{-1}(β,0)=\infty}$$

Please leave me feedback and corrections!

Hope for a sum representation:

In this question, a solution for

$$\int_0^1 Q^{-1}(x,x)dx=\int_0^1 Q^{-1}(x,x,Q(x,x)-x)dx$$

was found. I suppose we can use similar techniques. The goal is using the same series expansion in that question to find our goal integral of the two argument version because the three argument version does not have an easy series representation. It uses Big O notation:

$$Q^{-1}(x,x,x)= Q^{-1}(x,Q(x,x)-x) = ((1-Q(x,x)+x )x!)^\frac1x+\frac {((1-Q(x,x)+x )x!)^\frac2x}{x+1}+ \frac {(3x+5)((1-Q(x,x)+x )x!)^\frac3x}{2(x+1)^2(x+2)}+O\left((Q(x,x)-x-1)^\frac4x \right) $$

I hope that my solution can be proven true or false. One easy way to do this would be to get a rough approximation using the final answer from a truncated sum.

$\endgroup$
1
  • 2
    $\begingroup$ You may want to just use $Q_2$ instead of $\mathrm{\mathit{Q_2}}$. $\endgroup$
    – soupless
    Aug 25 at 12:21
7
$\begingroup$

To help with the possible answer:

The steps here are to use the series definition for the function using $C_{n,x}$ as the functional coefficients of the series, two Taylor series, a binomial theorem, and finally integrate. Each radius of convergence seems to be valid.


Please see the answer for:

On completing the solution for $\int_0^1 Q^{-1}(x,x) dx$.

which works based on the steps being valid. I will give links along the way as we find the answer. Unfortunately, it is almost impossible there is a closed form using known functions. I will also use information from the question itself. Let’s begin with this β constant:

$$ Q(β,β)=β, Q^{-1}(β,β)=β, Q^{-1}(β,β,β)=\infty, Γ(β,β)=β!,Γ(β)=β\,Γ(β,β), β= 0.259913533314486931564518…$$

Let’s start with the solution for $Q_2$. We can rewrite it into a form with 2 arguments using this identity. The goal is using this series expansion which already converges to about $\frac{3}{4}$ of the actual value of the integral at $3$ terms. Let $C_{n,x}$ be the coefficients of $((1-Q(x,x)+x )x!)^\frac nx$ in the expansion:

$$Q^{-1}(x,x,x)= Q^{-1}(x,Q(x,x)-x) = ((1-Q(x,x)+x )x!)^\frac1x+\frac {((1-Q(x,x)+x )x!)^\frac2x}{x+1}+ \frac {(3x+5)((1-Q(x,x)+x )x!)^\frac3x}{2(x+1)^2(x+2)}+O\left((Q(x,x)-x-1)^\frac4x \right) +…=((1-Q(x,x)+x )x!)^\frac1x C_{1,x}+ ((1-Q(x,x)+x )x!)^\frac2x C_{2,x}+ ((1-Q(x,x)+x )x!)^\frac3x C_{3,x}+… $$

$$Q_2\mathop=^\text{def}\int_0^β Q^{-1}(x,x,x)dx=\int_0^ β Q^{-1}(x,Q(x,x)-x)dx=\int_0^β\sum_{n=1}^\infty C_{n,x} ((1-Q(x,x)+x )x!)^\frac nx dx$$

The next step is to use the Taylor series of $x!^\frac nx$ at $x=0$ as the Taylor series for $(1-Q(x,x)+x )x!)^\frac nx$ is very complicated. We can choose an appropriate value at which to center the series to ensure convergence. Here is a graphical visualization. Note that this amazing expansion has the Euler-Mascheroni constant γ, Apéry’s Constant, the Zeta function, and uses Big-O Notation. Notice the polynomials of other constants, n, and x:

$$\sum_{m=0}^\infty \frac{\left(\frac{d^m}{dx^m}x!^\frac nx\right)_{x=0}\ x^m}{m!}= e^{- γ n}\left(1 +\frac{ π^2 n x}{12} + \frac{ n x^2 (π^4 n -96ζ(3))}{288} + \frac{π^2 n x^3 (5 π^4 n^2 + -1440 n ζ(3) + 144 π^2)}{51840}+ \frac{n x^4 (5 π^8 n^3 -2880 π^4 n^2 ζ(3) + 576 n (π^6 + 240ζ^2(3)) - 497644ζ(5)))}{2488320} + O\left(x^5\right)\right)$$

$$\int_0^β\sum_{n=1}^\infty C_{n,x} ((1-Q(x,x)+x )x!)^\frac nx dx= \int_0^β\sum_{n=1}^\infty C_{n,x} (1-Q(x,x)+x )^\frac nx \sum_{m=0}^\infty \frac{\left(\frac{d^m}{dx^m}x!^\frac nx\right)_{x=0}\ x^m}{m!} dx =\sum_{n=1}^\infty \sum_{m=0}^\infty \frac{\left(\frac{d^m}{dx^m}x!^\frac nx\right)_{x=0}}{m!} \int_0^β C_{n,x} x^m (1-Q(x,x)+x )^\frac nx dx$$

Now let’s use the property of β that $Q(β,β)=β$ to center the final needed Taylor series expansion about $x=β$ which also converges as shown here using approximations for β. The reason β is used is because the derivative contains many $Q(x,x)$ terms and coefficients which would just become β at $x= β$ as seen below.

Note that β!=Γ(β,β) and that this expansion uses the Meijer-G function, and Polygamma function. Notice the recurring pattern of similar functions and exponentiation which show the nth derivative would have a long, but solvable, series representation.I cannot think of a good way to integrate $(1-Q(x,x)+x )^\frac nx$ or something related in simpler terms, so this interesting Taylor series is the next best option.

If someone could simplify the Meijer G terms into a representation with less general functions, that would be very appreciated:

$$\sum_{k=0}^\infty \frac{\left(\frac{d^k}{dx^k} (1-Q(x,x)+x)^\frac nx\right)_{x=β}(x-β)^k}{k!}=1- n\left(\frac{\, G_{2, 3}^{3, 0} \left(β\big|\,^{\ 1, 1} _{0, 0, β}\right)}{β!} -\frac{β^{β - 1}}{β! e^{β}} +\ln(β) - ψ( β) - \frac1β \right)(x-β) +\frac12 \left[n \left(\ln(β)-\frac{G_{2, 3}^{3, 0} \left(β\big|\,^{\ 1, 1} _{0, 0, β}\right)}{β!} -\frac{e^{-β} β ^{β - 1}}{β!} - ψ( β) - \frac1β\right)^2 -\frac nβ \left(1-\frac{G_{2, 3}^{3, 0} \left(β\big|\,^{\ 1, 1} _{0, 0, β}\right)}{Γ(β)} + \frac{e^{-β} β^{β - 1}}{Γ(β)}- \ln(β) β + β ψ( β) \right) ^2 - \frac nβ \left(\ln(β)\left(\frac{G_{2, 3}^{3, 0} \left(β \big|\,^{\ 1, 1} _{0, 0, β}\right)}{Γ(β)}+ \frac{e^{-β} β ^{β - 1}}{Γ(β)} -\ln(β) β - β ψ(β)\right) + ψ( β)\left( \frac{G_{2, 3}^{3, 0} \left(β \big|\,^{\ 1, 1} _{0, 0, β}\right)}{Γ(β)} - \frac{e^{-β} β ^{β - 1}}{Γ(β)} + \ln(β) - ψ( β)\right)+ \frac{G_{1, 2}^{2, 0} \left(β \big|\,^{\quad 0}_{ -1,β-1} \right) - 2 G_{3, 4}^{4, 0} \left(β \big|\,^{\ 1, 1, 1} _{0, 0, 0, β}\right) - \ln(β) G_{2, 3}^{3, 0} \left(β \big|\,^{\ 1, 1} _{0, 0, β}\right)}{Γ(β)} + \frac{ψ( β) G_{2, 3}^{3, 0}\left(β \big|\,^{\ 1, 1}_ {0, 0, β}\right)}{Γ(β)} - \frac{e^{-β} β^{β - 1}}{Γ(β) }+ e^{-β}β ^{β - 1}\frac {1-\frac1 β + \ln(β))}{Γ(β)} - \frac{e^{-β} β ^{β - 1}ψ(β)}{Γ(β)} - \frac1β + β ψ^{(1)}( β)\right) - 2 n \left(\frac{e^{-β} β ^{β - 1}}{Γ(β)}-\frac{G_{2, 3}^{3, 0} \left(β \big|\,^{\ 1, 1} _{0, 0, β}\right)}{Γ(β)} - \frac{\ln(β)}{β} + \frac{ ψ(β)}{β} + 1\right)\right](x-β)^2+O\left(x^3\right)$$

Note that $$G_{2, 3}^{3, 0} \left(β \big|\,^{\ 1, 1} _{0, 0, β}\right)$$ cannot be written in terms of Regularized $_2\text F_2$ hypergeometric functions because the difference of the $0$s is an integer as seen in the link.

Also note that $$G_{1, 2}^{2, 0} \left(β \big|\,^{\quad 0}_{ -1,β-1} \right)=Γ(β)$$

Finally we can actually integrate. Note that I could have easily done a more general Taylor series, but these may have convergence problems and are much more complicated when written out. I try not to use just a single integrand Taylor Series as it gives no insight into the problem and can be used to evaluate any indefinite integral.

Problems, like this one, are more engaging when series expansions are simpler.

Note that the integral of $$\int_0^βC_{n,x} x^m (x-β)^k dx$$ does not have a closed form for the general nth term. However, we can use normal binomial theorem as the powers are summation indices and are natural numbers, so no convergence issues. Again, note that the choice of previous series expansions has consequences for further integration steps:

$$\sum_{n=1}^\infty \sum_{m=0}^\infty \frac{\left(\frac{d^m}{dx^m}x!^\frac nx\right)_{x=0}}{m!} \int_0^β C_{n,x} x^m (1-Q(x,x)+x )^\frac nx dx = \sum_{n=1}^\infty \sum_{m=0}^\infty \frac{\left(\frac{d^m}{dx^m}x!^\frac nx\right)_{x=0}}{m!} \sum_{k=0}^\infty \frac{\left(\frac{d^k}{dx^k} (1-Q(x,x)+x)^\frac nx\right)_{x=β}(x-β)^k}{k!} \int_0^β C_{n,x} x^m (x-β)^k dx = \sum_{n=1}^\infty \sum_{m=0}^\infty \frac{\left(\frac{d^m}{dx^m}x!^\frac nx\right)_{x=0}}{m!} \sum_{k=0}^\infty \frac{\left(\frac{d^k}{dx^k} (1-Q(x,x)+x)^\frac nx\right)_{x=β}(x-β)^k}{k!} \int_0^β C_{n,x} x^m (x-β)^k dx= \sum_{n=1}^\infty \sum_{m=0}^\infty \sum_{k=0}^\infty \sum_{t=0}^k \left(\frac{d^m}{dx^m}x!^\frac nx\right)_{x=0} \left(\frac{d^k}{dx^k} (1-Q(x,x)+x)^\frac nx\right)_{x=β} \frac{(-1)^{k+t}β^{k-t}}{m!t!(k-t)!}\int_0^β C_{n,x} x^{k+t} dx $$

Now let’s use the definition of $C_{n,x}$ to finish one of the answers. Note that I will not expand the summations or compact the expanded series for simplicity in the final answer. Note there are many possible simplifications. The integral now looks like an Incomplete Beta function or Lerch Transcendent function as a result of the series expansion. Note that I could have integrated up to 4 or more terms, but this becomes tedious and has more interesting complicated functions:

$$Q_2= \sum_{n=1}^\infty \sum_{m=0}^\infty \sum_{k=0}^\infty \sum_{t=0}^k \left(\frac{d^m}{dx^m}x!^\frac nx\right)_{x=0} \left(\frac{d^k}{dx^k} (1-Q(x,x)+x)^\frac nx\right)_{x=β} \frac{(-1)^{k+t}β^{k-t}}{m!t!(k-t)!}\int_0^β C_{n,x} x^{k+t} dx = \sum_{n=1}^\infty \sum_{m=0}^\infty \sum_{k=0}^\infty \sum_{t=0}^k \left(\frac{d^m}{dx^m}x!^\frac nx\right)_{x=0} \left(\frac{d^k}{dx^k} (1-Q(x,x)+x)^\frac nx\right)_{x=β} \frac{(-1)^{k+t}β^{k-t}}{m!t!(k-t)!}\int_0^β \left[1+\frac{1}{x+1}+\frac{3x+5}{2(x+1)^2(x+2)}+\frac{8x^2+33x+31}{3(x+1)^3(x+2)(x+3)}+…\right] x^{k+t} dx = \sum_{n=1}^\infty \sum_{m=0}^\infty \sum_{k=0}^\infty \sum_{t=0}^k \left(\frac{d^m}{dx^m}x!^\frac nx\right)_{x=0} \left(\frac{d^k}{dx^k} (1-Q(x,x)+x)^\frac nx\right)_{x=β} \frac{(-1)^{k+t}β^{k-t}}{m!t!(k-t)!}\left[\frac{β^{k+t+1}}{k+t+1}+(-1)^{k+t+1}B_{-β}(k+t+1,0) + β^{k + t + 1} \left(\frac14 \left(2 (β + 1) (1-2 k - 2 t )(k+t+1) Φ(-β,1,k+t+1) - (β+ 1)(k+t+1)Φ\left(-\fracβ2,1,k+t+1\right) \right) - k - t - 1\right)+…\right] =.221221…$$

Keeping the order of indices in mind, you can write

$$Q_2=\int_0^β Q^{-1}(x,x,x)dx=\int_0^β Q^{-1}(x, Q(x,x)-x)dx=\sum _{\mathop{\mathop{n\ge 1}\limits_{m,k\ge0}}\limits_{0\le t\le m}}\left(\frac{d^m}{dx^m}x!^\frac nx\right)_{x=0} \left(\frac{d^k}{dx^k} (1-Q(x,x)+x)^\frac nx\right)_{x=β} \frac{(-1)^{k+t}β^{k-t}}{m!t!(k-t)!}\int_0^β C_{n, x} x^{k+t} dx=\sum_{\mathop{\mathop{n\ge 1}\limits_{m,k\ge0}}\limits_{0\le t\le m}}\left(\frac{d^m}{dx^m}x!^\frac nx\right)_{x=0} \left(\frac{d^k}{dx^k} (1-Q(x,x)+x)^\frac nx\right)_{x=β} \frac{(-1)^{k+t}β^{k-t}}{m!t!(k-t)!}\left[\frac{β^{k+t+1}}{k+t+1}+(-1)^{k+t+1}B_{-β}(k+t+1,0) + β^{k + t + 1} \left(\frac14 \left(2 (β + 1) (1-2 k - 2 t )(k+t+1) Φ(-β,1,k+t+1) - (β+ 1)(k+t+1)Φ\left(-\fracβ2,1,k+t+1\right) \right) - k - t - 1\right)+…\right] $$

Note there may be a typo. I need to know if each step is valid and would make a true solution for $Q_2$. Any other simplifications, like an explicit formula for $C_{n,x}$, is appreciated. Please correct me and give me feedback!

$\endgroup$
1
  • 1
    $\begingroup$ Please help me know if this answer works; you can answer as well. Also what are simplifications for the answer if it works? $\endgroup$ Sep 27 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.