Dissecting a Wild Function: Derivative + Infinite Series + Infinite Product

We’re studying a complicated function

$$F(x) = \frac{d}{dx}\Bigg[ e^{x^{2}} \int_{0}^{x} \frac{\sin(t^{3}) + \ln(1+t^{2})}{1+t^{4}}\,dt \Bigg] - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \cos(n x^{2} + \sqrt{n}) - \lim_{N\to\infty} \prod_{k=1}^{N} \left(1+\frac{x^{2}}{k^{2}} e^{-x/k}\right).$$

The goal here is to understand what each piece means and how it behaves, not to compute an exact closed form.

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1. First term: derivative of a product with an integral

The first term is

$$G(x) = \frac{d}{dx}\Bigg[ e^{x^{2}} \int_{0}^{x} \frac{\sin(t^{3}) + \ln(1+t^{2})}{1+t^{4}}\,dt \Bigg].$$

Inside the derivative, we have a product:

• $e^{x^{2}}$: a rapidly growing smooth function.
• $I(x) = \int_{0}^{x} \dfrac{\sin(t^{3}) + \ln(1+t^{2})}{1+t^{4}}\,dt$: an integral whose upper limit is $x$, so it depends on $x$.

To differentiate $e^{x^{2}} I(x)$, use the product rule:

$$G(x) = (e^{x^{2}})' I(x) + e^{x^{2}} I'(x).$$

1. Derivative of $e^{x^{2}}$:

$$(e^{x^{2}})' = 2x e^{x^{2}}.$$

2. Derivative of the integral $I(x)$: by the Fundamental Theorem of Calculus, if $I(x) = \int_0^x f(t)\,dt,$ then $I'(x) = f(x).$ Here

$$f(t) = \frac{\sin(t^{3}) + \ln(1+t^{2})}{1+t^{4}},$$

so

$$I'(x) = \frac{\sin(x^{3}) + \ln(1+x^{2})}{1+x^{4}}.$$

Putting this together:

$$G(x) = 2x e^{x^{2}} I(x) + e^{x^{2}} \cdot \frac{\sin(x^{3}) + \ln(1+x^{2})}{1+x^{4}}.$$

So the first term is a smooth function built from:

• a growing factor $e^{x^{2}}$,
• an accumulated integral $I(x)$,
• plus a contribution from the instantaneous value of the integrand at $t = x$.

In the visualization, we’ll treat this term as a green-ish curve you can show or hide, and we approximate the integral numerically.

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2. Second term: an infinite oscillating series

The second term is

$$S(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \cos(n x^{2} + \sqrt{n}).$$

Features:

• Oscillation: each term is a cosine with phase $n x^{2} + \sqrt{n}$; varying both in $n$ and $x$.
• Alternating sign: $(-1)^n$ flips the sign of each term.
• Damping: the factor $\dfrac{1}{n^2}$ makes the terms shrink quickly, so the series converges for all real $x$.

Mathematically:

• For large $n$, $\bigg|\frac{(-1)^n}{n^2} \cos(n x^2 + \sqrt{n})\bigg| \le \frac{1}{n^2},$ and $\sum 1/n^2$ converges, so $S(x)$ is absolutely convergent.
• Being a nice sum of smooth terms, under standard theorems, $S(x)$ is smooth in $x$ and you can usually differentiate term-by-term on bounded intervals.

In the visualization, we truncate the sum at some $N$ (e.g., $N=50$ or user-controlled), and show how the partial sums approximate the series as a magenta curve.

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3. Third term: an infinite product

The third term is

$$P(x) = \lim_{N\to\infty} \prod_{k=1}^{N} \Bigg(1 + \frac{x^{2}}{k^{2}} e^{-x/k}\Bigg).$$

This is an infinite product: instead of adding infinitely many terms, we multiply infinitely many factors.

Each factor:

• $1 + \dfrac{x^{2}}{k^{2}} e^{-x/k}$.
• For fixed $x$, as $k\to\infty$, $x^2/k^2 \to 0$, so each new factor is very close to 1.

Convergence idea:

• For large $k$, $\dfrac{x^2}{k^2} e^{-x/k}$ behaves like $\dfrac{C}{k^2}$, so the series $\sum_k \dfrac{x^2}{k^2} e^{-x/k}$ converges.
• A classical criterion: if $\sum |a_k| < \infty$, then $\prod (1 + a_k)$ converges to a nonzero limit (assuming terms avoid hitting 0).
• Here, $a_k = \dfrac{x^2}{k^2} e^{-x/k}$, so the product converges for any fixed real $x$.

At $x = 0$, every factor becomes $1$, so

$$P(0) = 1.$$

In the visualization, we approximate the product up to some $N$, track the cumulative product, and plot an orange curve.

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4. Putting it together: the full function $F(x)$

We define

$$F(x) = G(x) - S(x) - P(x).$$

Conceptual behavior:

• Smoothness: each component (derivative-of-integral, convergent series, convergent product) is smooth in $x$. So $F(x)$ is very smooth.
• Growth / decay:
  • $G(x)$ can grow quickly because of $e^{x^2}$.
  • $S(x)$ stays bounded, oscillating.
  • $P(x)$ is positive and typically not enormous because each factor is close to 1.

At $x=0$:

1. Integral term:
   • $I(0) = \int_0^0 (\cdots) dt = 0$.
   • So the product inside the derivative is $e^{0} \cdot 0 = 0$.
   • But we care about $G(0)$, the derivative at 0:

$$G(0) = 2\cdot 0 \cdot e^{0} I(0) + e^{0} \cdot \frac{\sin(0^3) + \ln(1+0^2)}{1+0^4} = 0 + 1 \cdot \frac{0 + \ln 1}{1} = 0.$$

2. Series term:

$$S(0) = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \cos(\sqrt{n}).$$

This has no simple closed form, but it converges to some finite real number.

3. Product term:

$$P(0) = 1.$$

So

$$F(0) = G(0) - S(0) - P(0) = 0 - S(0) - 1 = -S(0) - 1.$$

In other words, $F(0)$ is a specific real constant, determined by that convergent series minus 1.

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5. What the visualization shows

The interactive canvas helps you build intuition by:

• Plotting on a horizontal $x$-axis (with a Cartesian transform) the following curves:

  • First term $G(x)$ (cyan/green): derivative of the integral-product.
  • Series $S(x)$ (pink/magenta): partial sums with $N$ terms.
  • Product $P(x)$ (yellow/orange): partial products with $N$ factors.
  • Full function $F(x)$ (bright cyan): combination of the three.

• Letting you control:

  • the x–range so you can zoom in/out,
  • the number of terms used to approximate the series and product,
  • which components are visible.

• Using light animation to show how partial sums/products stabilise as $N$ increases.

Focus points while exploring:

• Near $x=0$: observe that all components behave smoothly and that the product tends to 1.
• As $|x|$ grows: watch the first term grow due to $e^{x^2}$, while the series still oscillates and the product deforms more gently.
• Vary the term count: see partial sums/products converge visually.

This doesn’t give a closed-form formula for $F(x)$, but it gives a strong qualitative picture of how each component behaves and how they interact to form the full function.