Minimum Value of Definite Integral Function with Inverse Tan Upper Limit

To find the minimum value of the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \int_{0}^{x \tan^{-1} x} \frac{e^{(t - \cos t)}}{1 + t^{2023}} \, dt$

we first analyze the upper limit of the definite integral, let $u(x) = x \tan^{-1} x$.

1. Analysis of the upper limit $u(x)$:

   • If $x > 0$, then $\tan^{-1} x > 0$, which gives $u(x) = x \tan^{-1} x > 0$.
   • If $x < 0$, then $\tan^{-1} x < 0$, which also gives $u(x) = x \tan^{-1} x > 0$.
   • If $x = 0$, then $u(0) = 0 \cdot \tan^{-1}(0) = 0$.

   Thus, for all $x \in \mathbb{R}$, we have $u(x) \ge 0$, with $u(x) = 0$ if and only if $x = 0$.

2. Analysis of the integrand: Let the integrand be $h(t) = \frac{e^{t - \cos t}}{1 + t^{2023}}$.

   • The exponential term $e^{t - \cos t} > 0$ for all $t \in \mathbb{R}$.
   • For $t \ge 0$, we have $t^{2023} \ge 0$, which implies $1 + t^{2023} \ge 1 > 0$.

   Therefore, for all $t \ge 0$, the integrand $h(t) > 0$.

3. Analysis of $f(x)$:

   • For any $x \neq 0$, the upper limit $u(x) > 0$. Since $h(t) > 0$ on the interval $[0, u(x)]$, the integral of a strictly positive function over a non-zero positive interval is strictly positive: $f(x) = \int_{0}^{u(x)} h(t) \, dt > 0 \quad \text{for all } x \neq 0$
   • For $x = 0$, the lower and upper limits of integration are identical: $f(0) = \int_{0}^{0} h(t) \, dt = 0$

4. Derivative Analysis (Verification): Using Leibniz's Rule of Differentiation under the integral sign: $f'(x) = h(u(x)) \cdot u'(x)$ $f'(x) = \left( \frac{e^{x \tan^{-1} x - \cos(x \tan^{-1} x)}}{1 + (x \tan^{-1} x)^{2023}} \right) \cdot \left( \tan^{-1} x + \frac{x}{1 + x^2} \right)$

   Since $h(u(x)) > 0$ for all $x \in \mathbb{R}$:

   • For $x < 0$, $u'(x) < 0 \implies f'(x) < 0$ (the function $f(x)$ is strictly decreasing).
   • For $x > 0$, $u'(x) > 0 \implies f'(x) > 0$ (the function $f(x)$ is strictly increasing).
   • At $x = 0$, $f'(0) = 0$.

This confirms that $x = 0$ is the unique absolute minimum point of $f(x)$.

The minimum value of $f(x)$ is $f(0) = 0$.