Are there series/non-trivial integrals that converge to $e\pi$?

Question

I've tried for hours but I can't really find series representations without explicitly involving either $e$ or $\pi$ within the series expression.

Edit: I am asking this for $e$ times $\pi$, and not for $e$ or $\pi$ like some edits have suggested.

Answer 1

Here's a somewhat unpleasant answer, but an answer nonetheless.

$$ e^x = \sum_{n=0}^{+\infty} \frac{x^n}{n!} \qquad \text{and} \qquad \arcsin x = \sum_{n=0}^{+\infty} \frac{{2n \choose n} x^{2n+1}}{4^n(2n+1)}$$

Let $x = 1$ in both series, and in the second series make sure to scale appropriately ($\arcsin 1 = \pi/2$, so $\pi = 2\arcsin 1 = \cdots$).

Now use the Cauchy product to multiply the two series representations.

Answer 2

Here is one strategy to construct integrals converting to $\pi e$ relatively easily. Basically all we need is a differentiable function $F(t)$ (which does not contain $\pi$ or $e$ in itself) such that $\lim_{t \to t_0}F(t) = \pi e$. Then supposing $F'(x)=f(x)$ we have $$ \int_0^{t_0}f(x) \,dx = \lim_{t\to t_0} \int_0^{t}f(x) \,dx = \lim_{t\to t_0} F(t) = \pi e $$

So we just need to find suitable function with limit we want, without containing $\pi$ or $e$ in the function itself. It seems also desirable to find function which has simple derivative, and that is the hardest part... So let's construct one and perhaps better one will come up later. Let's try

$$ F(x)=2 \arcsin x\cdot (2-x)^{\frac1{1-x}} $$

It is not hard to verify that $\lim_{t\to 1} F(t) = \pi e$ and so we have $$ \int_0^{t_0}F'(x) \,dx = \int_0^{1}\frac{2(2-x)^\frac1{1-x}}{\sqrt{1-x^2}}+2 \arcsin x (2-x)^\frac1{1-x}\left(\frac{\ln (2-x)}{(1-x)^2}-\frac1{(1-x)(2-x)}\right)\,dx = \pi e $$ Even though the end result is not very satisfying, it is in principle possible to try other choices for $F(x)$ and try to find something which will look better.

Edit: One can do the similar also for infinite series, we need sequence $a_n$ converging to $\pi e$ with $a_1=0$. Then we have $$\sum_{n=1}^{\infty} a_n -a_{n-1} = \lim_{k\to \infty} \sum_{n=1}^{k} a_n-a_{n-1} = \lim_{k\to \infty} a_k-a_{0} = \pi e$$ so choosing similarly as above let $a_n = 2\arcsin\left(\frac{n}{n+1}\right)\left(1+\frac{1}{n}\right)^n$ for $n>0$ and $a_0=0$, then $$\sum_{n=1}^{\infty} 2\arcsin\left(\frac{n}{n+1}\right)\left(1+\frac{1}{n}\right)^n -2\arcsin\left(\frac{n-1}{n}\right)\left(1+\frac{1}{n-1}\right)^{n-1} = \pi e$$