da Mephlip » 27/09/2020, 17:48
Testo nascosto, fai click qui per vederlo
For (1): I believe that there is a typo in the right hand side of $I(0)$, specifically a factor $\frac{1}{2}$ is missing since from the definition of $I(p)$ it follows that
$$I(0)=\int_0^{\infty} \frac{1}{2(1+x^2)} \text{d}x$$
We have that
$$I(0)=\int_0^{\infty} \frac{1}{2(1+x^2)} \text{d}x=\frac{1}{2} \cdot \frac{\pi}{2}=\frac{\pi}{4}$$
For the evaluation of $I(1)$ and $I(2)$ we will use a technique that can be useful in these kind of integrals; this technique will also give the idea for the general case. The idea is to notice that the interval of integration is invariant under the map $x\mapsto \frac{1}{x}$, hence
$$I(1)=\int_0^{\infty} \frac{1}{(1+x^2)(1+x)}\text{d}x=\int_0^{\infty} \frac{1}{x^2\left(1+\frac{1}{x^2}\right)\left(1+\frac{1}{x}\right)} \text{d}x=\int_0^{\infty} \frac{x}{(1+x^2)(1+x)}\text{d}x=$$
$$=\int_0^{\infty} \frac{1}{1+x^2} \text{d}x-I(1) \Rightarrow 2I(1)=\int_0^{\infty} \frac{1}{1+x^2} \text{d}x=\frac{\pi}{2}\Rightarrow I(1)=\frac{\pi}{4}$$
And
$$I(2)=\int_0^{\infty} \frac{1}{(1+x^2)^2}\text{d}x=\int_0^{\infty} \frac{1}{x^2\left(1+\frac{1}{x^2}\right)^2} \text{d}x=\int_0^{\infty} \frac{x^2}{(1+x^2)^2}\text{d}x=$$
$$=\int_0^{\infty} \frac{1}{1+x^2} \text{d}x -I(2) \Rightarrow 2I(2)=\int_0^{\infty} \frac{1}{1+x^2} \text{d}x=\frac{\pi}{2} \Rightarrow I(2)=\frac{\pi}{4}$$
For (2): By letting $x \mapsto \frac{1}{x}$, it follows that
$$I(p)=\int_0^{\infty} \frac{1}{(1+x^2)(1+x^p)} \text{d}x=\int_0^{\infty} \frac{x^p}{(1+x^2)(1+x^p)} \text{d}x =\int_0^{\infty} \frac{1}{1+x^2}\text{d}x-I(p) \Rightarrow$$
$$\Rightarrow 2I(p)= \int_0^{\infty} \frac{1}{1+x^2} \text{d}x=\frac{\pi}{2} \Rightarrow I(p)=\frac{\pi}{4}$$
For all $p\in\mathbb{R}$.
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