Evaluating the Series Involving arctan Functions for Large x

In this article, we examine a specific series involving arctan functions and provide an in-depth analysis of how to evaluate it for large values of x. The series in question is:

$$displaystyle -S_x -sum_{n1}^infty arctanleft(frac{x}{1-x-n^2}right) sum_{n1}^infty arctanleft(frac{x}{x-n^2-1}right) tag{1}$$

To facilitate this evaluation, we break the sum into three parts, each of which we analyze in detail.

Splittng the Sum into Three Parts

Let's denote the integer part of a number as [x]. We can split the sum (1) into the following three parts:

$$ displaystyle -S_x S_1 S_2 S_3 sum_{n1}^{[x]-2} arctanleft(frac{x}{x-n^2-1}right) sum_{n[x]-2}^infty arctanleft(frac{x}{x-n^2-1}right) arctanleft(frac{x}{x-[x]^2-1}right) arctanleft(frac{x}{x-[x]-1^2-1}right) arctanleft(frac{x}{x-[x]-1^2-1}right) tag{2} $$

The last three terms represent S_3.

Evaluating S_3

The evaluation of S_3 is straightforward. For large x, we have:

$$ displaystyle arctanleft(frac{x}{x-[x]^2-1}right) -left(frac{pi}{2} - frac{1-x-[x]^2}{x}right) O(1) tag{3} $$

Therefore, S_3 is of order O(1).

Evaluating S_1 and S_2 Using Euler-Maclaurin Formula

To evaluate S_1 and S_2, we use the Euler-Maclaurin formula:

$$ displaystyle sum_{na}^b f(n) int_a^b f(n) , d n frac{1}{2} left[f(b) f(a)right] frac{1}{12} left[f'(b) - f'(a)right] ... tag{4} $$

For example, for S_1, we get:

$$ displaystyle sum_{n1}^{[x]-2} arctanleft(frac{x}{x-n^2-1}right) int_{n1}^{[x]-2} arctanleft(frac{x}{x-n^2-1}right) , d n left[frac{1}{2} left(arctanleft(frac{x}{x-[x]-2^2-1}right) - arctanleft(frac{x}{x-1^2-1}right)right)right] left[frac{1}{12} left(arctanleft(frac{2x([x]-2)-x}{left(x-[x]-2^2-1right)^2 x^2}right) - arctanleft(frac{2x x-1}{left(x-1^2-1right)^2 x^2}right)right)right] ... tag{5} $$

We see that the second term gives the contribution of order O(1/x), and the third term gives the contribution of order O(1/x^2), and so on. The same story applies to S_3.

Final Integration and Approximation

With the same accuracy, we can write:

$$ displaystyle -S_x int_1^infty arctanleft(frac{x}{n-x^2-1}right) , d n int_0^infty arctanleft(frac{x}{n-x^2}right) , d n tag{6} $$

To integrate (6) by parts, make the change n - x t:

$$ displaystyle -S_x sim 2x^2 int_{-x}^infty frac{t, dt}{x^2 t^4} - 2x int_{-infty}^infty frac{t^2, dt}{x^2 t^4} tag{7} $$

After further simplification, we get:

$$ displaystyle -S_x sim x^4 int_1^infty frac{dt}{x^2 x^4 t^2} 2x int_{-infty}^infty frac{t^2, dt}{x^2 t^4} - 2x^4 int_1^infty frac{t^2, dt}{x^2 x^4 t^4} tag{8} $$

The first and third terms in (8) give a contribution that is of order O(1). The second term diverges as x approaches infinity. Keeping only the leading asymptotic term:

$$ displaystyle -S_x sim 2x int_{-infty}^infty frac{t^2, dt}{x^2 t^4} 2sqrt{x} int_{-infty}^infty frac{t^2, dt}{1 t^4} sqrt{x} int_0^1 left(1 - s^{-1/4}right) s^{-3/4} , ds tag{9} $$

The boxed result is:

$$ displaystyle boxed{-S_x sim sqrt{x} Bleft(frac{1}{4}, frac{3}{4}right) sqrt{2pi} sqrt{x} quad text{as} quad x to infty} tag{10} $$

Verification with WolframAlpha

Let's verify that our asymptotic result works. Using WolframAlpha:

$$ x 100:quad -S_{100} 37.36 sqrt{2pi} sqrt{100} 44.43 $$

$$ x 300:quad -S_{300} 76.08 sqrt{2pi} sqrt{300} 76.95 $$

$$ x 500:quad -S_{500} 98.44 sqrt{2pi} sqrt{500} 99.34 $$

$$ x 1000:quad -S_{1000} 139.6 sqrt{2pi} sqrt{1000} 140.5 $$