反常积分收敛性证明求助：证明∫₀^∞ dx/(1+x⁴) ≤ 4/3

Hey there! Great question—splitting the integral into two parts is exactly the right move here, since comparing to $\frac{1}{x^4}$ over the entire interval doesn’t work (as you noticed, it diverges near 0). Let’s break this down step by step:

Step 1: Split the integral at x=1 #

We can rewrite the original integral as the sum of two separate integrals, which lets us handle each interval with a suitable comparison function:
$$\int_{0}^{{\infty}\frac{dx}{1+x}4} = \int_{0}^{{1}\frac{dx}{1+x}4} + \int_{1}^{{\infty}\frac{dx}{1+x}4}$$

Step 2: Bound the integral from 0 to 1 #

For $x \in [0,1]$, $x^4$ ranges from 0 to 1, so $1+x^4 \geq 1$. Taking reciprocals (since both sides are positive, the inequality direction stays the same):
$$\frac{1}{1+x^4} \leq 1$$
Integrating both sides from 0 to 1 gives:
$$\int_{0}^{{1}\frac{dx}{1+x}4} \leq \int_{0}^{1}1\ dx = 1$$

Step 3: Bound the integral from 1 to infinity #

For $x \in [1, \infty)$, $x^4 \geq x^4$ (obviously!), so $1+x^4 \geq x^4$. Taking reciprocals here reverses the inequality (since both denominators are positive):
$$\frac{1}{1+x^4} \leq \frac{1}{x^4}$$
Now compute the convergent integral of $\frac{1}{x^4}$ from 1 to infinity:
$$\int_{1}^{{\infty}\frac{dx}{x}4} = \lim_{b \to \infty} \int_{1}^{b}x{-4}\ dx = \lim_{b \to \infty} \left[ -\frac{1}{3x^3} \right]{1}^{b} = 0 - \left(-\frac{1}{3}\right) = \frac{1}{3}$$
This means:
$$\int{1}^{{\infty}\frac{dx}{1+x}4} \leq \frac{1}{3}$$

Step 4: Combine the results #

Adding the two bounds together gives exactly what we need to prove:
$$\int_{0}^{{\infty}\frac{dx}{1+x}4} = \int_{0}^{{1}\frac{dx}{1+x}4} + \int_{1}^{{\infty}\frac{dx}{1+x}4} \leq 1 + \frac{1}{3} = \frac{4}{3}$$

The core idea here is recognizing that the integrand behaves differently near 0 and at infinity—splitting the integral lets us use appropriate comparison functions for each region, avoiding the divergence problem you hit initially.

备注：内容来源于stack exchange，提问作者Garlic6522