For $$x \in \mathbb{R}$$, let $$\tan^{-1}(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$. Then the minimum value of the function $$f : \mathbb{R} \to \mathbb{R}$$ defined by $$f(x) = \int_0^{x\tan^{-1}x} \frac{e^{(t - \cos t)}}{1 + t^{2023}}\,dt$$ is

For every real number $$x$$ the function is defined as
$$f(x)=\int_{0}^{x\,\tan^{-1}x}\dfrac{e^{\,t-\cos t}}{1+t^{2023}}\;dt\qquad\qquad -(1)$$

Step 1 : Sign of the upper limit
Put $$\alpha(x)=x\,\tan^{-1}x$$.
  • If $$x\ge 0$$, then $$\tan^{-1}x\ge 0$$, so $$\alpha(x)\ge 0$$.
  • If $$x\le 0$$, then $$\tan^{-1}x\le 0$$; the product of two non-positive numbers is non-negative, so again $$\alpha(x)\ge 0$$.
Hence for every $$x\in\mathbb{R}$$ we have

$$\boxed{\;\alpha(x)=x\,\tan^{-1}x\;\ge 0\;}$$

Equality occurs only at $$x=0$$ because $$x=0$$ gives $$\alpha(0)=0$$, while for any $$x\ne 0$$ the product of two numbers of the same sign is positive.

Step 2 : Sign of the integrand on $$[0,\alpha(x)]$$
For $$t\ge 0$$ we examine each factor of the integrand
$$I(t)=\dfrac{e^{\,t-\cos t}}{1+t^{2023}}.$$
(i) Numerator: $$e^{\,t-\cos t}\gt 0$$ for all real $$t$$.
(ii) Denominator: when $$t\ge 0$$ the odd power $$t^{2023}\ge 0$$, so $$1+t^{2023}\ge 1\gt 0$$.
Therefore for every $$t\in[0,\alpha(x)]$$

$$\boxed{\;I(t)\gt 0\;}$$

Step 3 : Sign of the integral $$f(x)$$
Because the upper limit $$\alpha(x)$$ is non-negative, the integral in $$(1)$$ is taken over an interval on which the integrand is strictly positive (except possibly at $$t=0$$ where it is still non-negative). Hence

$$f(x)=\int_{0}^{\alpha(x)} I(t)\,dt\;\ge\;0 \qquad\text{for all }x\in\mathbb{R}.$$

Step 4 : Value at $$x=0$$
At $$x=0$$ the upper limit vanishes, so

$$f(0)=\int_{0}^{0}\dfrac{e^{\,t-\cos t}}{1+t^{2023}}\,dt=0.$$

Step 5 : Minimum value
We have shown
• $$f(x)\ge 0$$ for every $$x\in\mathbb{R}$$, and
• $$f(0)=0$$.
Therefore the smallest possible value of $$f(x)$$ is attained at $$x=0$$ and equals $$0$$.

Hence the minimum value of the function is 0.