Prove or disprove following inequality of integrals

Question

If $p$, $q$ are positive numbers such that $p<q$, prove that

$$\int_{0}^{\infty}\frac{dx}{1+x^{p}}> \int_{0}^{\infty}\frac{dx}{1+x^{q}}.$$

Answer 1

It is well known that

$$\int_{0}^{\infty}\frac{dx}{1+x^p}=\dfrac{\pi}{p \sin\tfrac{\pi}{p}}$$

(see for example a more general formula in https://math.stackexchange.com/q/48741)

Thus your question can be rephrased into :

$$1<p<q \ \implies \ \dfrac{\pi}{p \sin\tfrac{\pi}{p}}>\dfrac{\pi}{q \sin\tfrac{\pi}{q}}\tag{1}$$

Otherwise said :

$$\tfrac{\pi}{q}<\tfrac{\pi}{p}<\pi \ \implies \frac{\sin\tfrac{\pi}{p}}{\tfrac{\pi}{p}}<\frac{\sin\tfrac{\pi}{q}}{\tfrac{\pi}{q}} \tag{2} $$

itself equivalent to :

$$x_1<x_2<\pi \ \implies \frac{\sin x_2}{x_2}<\frac{\sin x_1}{x_1} \tag{3} $$

(3) being true because it means that function "cardinal sine" defined by :

$$\text{sinc}(x):=\frac{\sin x}{x}$$

is decreasing on interval $(0,\pi)$ (see figure below), a fact that can be verified in a straightforward manner.

Fig. 1 : On $(0,\pi)$ (red part of the curve), function "sinc" is decreasing.