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Conjecture: $$\sum_{r=0}^{\infty}\left(\frac{x^{2r}\left(r+1\right)^{x}}{r!}\right)\overset{?}{=}{}\frac{\Gamma\left(x+2\right)}{\pi x^{2}}\int_{0}^{\pi}e^{x^{2}\cos\left(\sin t\right)e^{\cos t}}\cos\left(x^{2}e^{\cos t}\sin\left(\sin t\right)-\left(x+1\right)t\right)dt$$

I am not able to confirm this numerically. Could anyone help.

Let, $$I(x)=\int_{0}^{\pi}e^{x^{2}\cos\left(\sin t\right)e^{\cos t}}\cos\left(x^{2}e^{\cos t}\sin\left(\sin t\right)-\left(x+1\right)t\right)dt$$ $$\frac{I(n)(n+1)!}{\pi n^2e^{n^2}} \in \text{N}\ \forall \ n\in \text{N}$$ The Integral can be broken down as follows: $$I(x)=I_1(x)+I_2(x)$$ $$I_1(x)=\int_{0}^{\pi}e^{x^{2}\cos\left(\sin t\right)e^{\cos t}}\cos\left(x^{2}\sin\left(\sin t\right)e^{\cos t}\right)\cos\left(\left(x+1\right)t\right)dt$$ $$I_2(x)=\int_{0}^{\pi}e^{x^{2}\cos\left(\sin t\right)e^{\cos t}}\sin\left(x^{2}\sin\left(\sin t\right)e^{\cos t}\right)\sin\left(\left(x+1\right)t\right)dt$$

For small values $I_1(x)=I_2(x)$, I am not able to confirm the equality.
The Integral can be broken down more after expanding the Trigonometric Terms.

This is based on my previous question asked here. Summation $\sum_{n=0}^{\infty}\left(\frac{x^{2n}\left(n+1\right)^{x}\ }{n!}\right)$
If anyone could numerically verify whether the conjecture is true or not.
And is it possible to evaluate the Integral further?

Miracle Invoker
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1 Answers1

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Two interesting problems in a couple of days.

Numerically, the integrals make a lot of problems as soon as $x>4$.

Taking into account your previous remarks, what I computed is $$J_1(x)=e^{-x^2}\,I_1(x) \qquad \text{and} \qquad J_2(x)=e^{-x^2}\,I_2(x)$$

There is something which is interesting : for a WorkingPrecision=500

$$J_1(1)=J_2(1)=\frac \pi 2$$ which already seems to invalidate the conjecture.

Using WorkingPrecision=50 and truncating $$\left( \begin{array}{ccc} x & J_1(x) & J_2(x) \\ 1 & 1.5707963267948966192 & 1.5707963267948966192 \\ 2 & 30.368728984701334638 & 30.368728984701334638 \\ 3 & 753.39318823900229100 & 753.39318823900229100 \\ 4 & 23695.357869945895841 & 23695.357869945895841 \\ 5 & 910715.58530913653920 & 910715.58530913653920 \\ 6 & 41496402.422500335020 & 41496402.422500335020 \\ \end{array} \right)$$ which do not look like the sequence $$\{2,29,1279,113137,16697651,3698440237\}$$

Claude Leibovici
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  • I think I have another form for the sequence: $$\left[\frac{\text{d}^{n+1}}{\text{d}x^{n+1}}e^{n^2e^x}\right]_{x=0}\frac{1}{n^2e^{n^2}}$$ All of this seems related to Touchard Polynomials, but I am still in school and these seem out of my scope. Also thank you for the numerical evaluation, I will check up on that Integral again. My original question was to find the closed form of the function, but it seems to be difficult. – Miracle Invoker Aug 19 '23 at 04:26
  • $$I(n)=\int_{0}^{\pi}e^{n^{2}e^{\cos t}\cos\left(\sin t\right)}\cos\left(n^{2}e^{\cos t}\sin\left(\sin t\right)-\left(n+1\right)t\right)dt$$ $$S_n=\frac{(n+1)!}{\pi n^2e^{n^2}}I(n)$$

    It does seem to define the Sequence, but again I am using Desmos for numerical verification and it works for $n=1...3$, after that it cannot really compute clearly.

     – Miracle Invoker Aug 19 '23 at 04:52
  • @BlackEmperor. Why don't you register for Wolfram Cloud ? It is free. – Claude Leibovici Aug 19 '23 at 05:47
  • Oh! If it is free then I shall do so. Thanks. – Miracle Invoker Aug 19 '23 at 05:50
  • What you wrote for $S_n$ is strictely exact. The "small" problem is $I(n)$. Nice problem – Claude Leibovici Aug 19 '23 at 05:58
  • Now that I searched the Integral on Approach0, it seems to be the Integral Representation of Bell Numbers – Miracle Invoker Aug 19 '23 at 06:00