A family of improper integrals

[gugo82]

Problem:

1. Compute:
\[
\begin{split}
I(0) &:= \intop_0^\infty \frac{1}{1+x^2}\ \text{d} x \\
I(1) &:= \intop_0^\infty \frac{1}{(1+x^2)(1+x)}\ \text{d} x \\
I(2) &:= \intop_0^\infty \frac{1}{(1+x^2)^2}\ \text{d} x \; ,
\end{split}
\]
showing that these integrals share the same numerical value.

2. Prove that each integral:
\[
I(p) := \intop_0^\infty \frac{1}{(1+x^2)(1+x^p)}\ \text{d} x
\]
with $p in RR$ evaluates as the three integrals above.

[gugo82]

@ Mephlip: Yep! It's a typo.

[Mephlip]

I didn't think about about that, nice approach! I'm missing something or shouldn't it be

$$I(0)=\int_0^{\infty} \frac{1}{(1+x^2)(1+x^0)} \text{d}x=\int_0^{\infty} \frac{1}{(1+x^2)\cdot2} \text{d}x=\frac{\pi}{4}$$
Instead of $\frac{\pi}{2}$?

[gugo82]

Ok!

Here's another way:

By definition, $I(p)$ is a smooth function of $p$, hence differentiation under integral sign (w.r.t. $p$) works and gives:
\[
\begin{split}
I^\prime (p) &= - \intop_0^\infty \frac{x^p \log x}{(1+x^2)(1+x^p)^2}\ \text{d} x \\
&= - \intop_0^1 \frac{x^p \log x}{(1+x^2)(1+x^p)^2}\ \text{d} x - \intop_1^\infty \frac{x^p \log x}{(1+x^2)(1+x^p)^2}\ \text{d} x \\
&= - \intop_0^1 \frac{x^p \log x}{(1+x^2)(1+x^p)^2}\ \text{d} x - \intop_1^0 \frac{y^{-p} \log y^{-1}}{(1+y^{-2})(1+y^{-p})^2}\ \left(-\frac{1}{y^2}\right)\ \text{d} y \\
&= - \intop_0^1 \frac{x^p \log x}{(1+x^2)(1+x^p)^2}\ \text{d} x + \intop_0^1 \frac{y^p \log y}{(y^2+1)(y^p+1)^2}\ \text{d} y \\
&=0\;.
\end{split}
\]
Therefore $I(p)$ is a constant and equals $I(0) = pi/4$.

[Mephlip]

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 $p\in\mathbb{R}$.