Integration of $\int_{0}^{\infty}{\frac{1}{(x^2+1)(x^{2019}+1)}dx}$

Question

The original problem is:

The proof of the equation show above is what I want. I found that the first integral equals $\pi/4$. The changes of 2019 from 1 to any real numbers bigger than 1 cause no change of the final integration.

Answer 1

A numeric verification

Clear["Global`*"]

The integrand of the lhs can be written as 1/(x^2 + 1) - 1/((x^2 + 1)(x^2019 + 1))

x^2019/((x^2 + 1) (x^2019 + 1)) == 
  (1+x^2019)/((x^2+1)(x^2019+1)) - 1/((x^2+1)(x^2019+1)) == 
  1/(x^2 + 1) - 1/((x^2 + 1) (x^2019 + 1)) // Simplify

(* True *)

lhs1 = Integrate[1/(x^2 + 1), {x, 0, ∞}]

(* π/2 *)

The numeric part of the verification:

lhs2 = π ((NIntegrate[1/((x^2+1)(x^2019 + 1)), {x, 0, ∞}]/π) //
     RootApproximant)

(* π/4 *)

lhs = lhs1 - lhs2

(* π/4 *)

rhs = 1/2 Integrate[1/(x^2 + 1), {x, 0, ∞}]

(* π/4 *)

lhs == rhs

(* True *)

Answer 2

You should change the variable $x\to \frac{1}{x}$

(1/(1 + x^2)*1/(1 + x^2019) /. x -> 1/x)*D[x /. x -> 1/x, x] // Simplify

$$-\frac{x^{2019}}{\left(x^2+1\right) \left(x^{2019}+1\right)}$$

That is \begin{align} &\int_{0}^{\infty} \frac{1}{(1+x^2)(1+x^{2019})}\,\mathrm{d}x\\ =&\int_{\infty}^{0} \frac{1}{(1+(\frac{1}{x})^2)(1+(\frac{1}{x})^{2019})}\,\mathrm{d}\frac{1}{x}\\ =&\int_{\infty}^0-\frac{x^{2019}}{\left(x^2+1\right) \left(x^{2019}+1\right)}\,\mathrm{d}x \end{align}

So we get $$\int_{0}^{\infty} \frac{1}{(1+x^2)(1+x^{2019})}\,\mathrm{d}x=\int_{0}^{\infty}\frac{x^{2019}}{\left(x^2+1\right) \left(x^{2019}+1\right)}\,\mathrm{d}x$$

It means that all of them equal to $$\frac{1}{2}\int_{0}^{\infty} \frac{1}{(1+x^2)(1+x^{2019})}+\frac{x^{2019}}{\left(x^2+1\right) \left(x^{2019}+1\right)}\,\mathrm{d}x$$

Just equal to $$\frac{1}{2}\int_0^{\infty}\frac{1}{1+x^2}\,\mathrm{d}x=\frac{\pi}{4}$$

1/2 Integrate[1/(1 + x^2), {x, 0, ∞}]

π/4

Answer 3

Two chances:

(A)

(x^2019/((1 + x^2) (1 + x^2019)) - 
   1/2 (1 + x^2019)/((1 + x^2) (1 + x^2019))) // FullSimplify

(* (-1 + x^2019)/(2 (1 + x^2) (1 + x^2019)) *)

(B)

x^2019/((1 + x^2) (1 + x^2019)) - 
 1/2 (1 + x^2019)/((1 + x^2) (1 + x^2019))

(* -(1/(2 (1 + x^2))) + x^2019/((1 + x^2) (1 + x^2019)) *)

The [NIntegrate] results must be wrong too. We know for sure from NIntegrate that

NIntegrate[1/(1 + x^2), {x, 0, \[Infinity]}]

$\frac{\pi}{4}$

By the famous majorant criteria for infinite integrals we have ease to estimate that this has to be bigger then the

Take as a reference for example this page: Integral test for convergence

To do a harder attempt:

NIntegrate[1/(1 + x^2)*1/(1 + x^2019), {x, 0, 1}, 
 PrecisionGoal -> 100, WorkingPrecision -> 200]

(*

0.78522650738967301568031546551813653513306507248637888350236342703381\
7598418491157836764104246377397390853048300788950931805991775817967566\
46995443684292857095545206581998393114079894683526201949452667

*)

That fails and is indeed as expected smaller than $\frac{\pi}{4}$.

Plot[{1/(1 + x^2) , 1/((1 + x^2) (1 + x^2019)), 
  1/((1 + x^2) (1 + x^2020))}, {x, 0.8, 1.2}, 
 PlotLegends -> "Expressions", PlotRange -> {{0.8, 1.2}, Automatic}]

Just for value a little bigger than $1$ the second factor in the denominator explodes to very high values and the curves are really different.

Table[{1/(1 + x^2) , 1/((1 + x^2) (1 + x^2019)), 
   1/((1 + x^2) (1 + x^2020))}, {x, 0.5, 1.5, 0.5}] // TableForm

NIntegrate[1/((1 + x^2) (1 + x^2019)), {x, 0, 1}]

(* 0.785398 *) is a good approximation to the integral with the high requirements by [NIntegrate].

Despite it looks like a sharp jump it is not.

[NIntegrate] is much like Plot it needs the hand of an experienced Mathematician to give good results.

A powerful plot is for the answer of Your question:

Plot[NIntegrate[1/(1 + x^2)*1/(1 + x^2019), {x, 0, y}, 
  PrecisionGoal -> 100, WorkingPrecision -> 200], {y, 0, 1.05}, 
 PlotLegends -> "Expressions", PlotRange -> Full]

So the two calculations at the beginning of my answer did prove already the equation does not hold. The integral value is close to $0.785398$. The functions series $f(x,n)=\frac{1}{(1+x^2)(1+x^n))}$ with $n$ Integers is strongly monoton dropping with increasing $n$. $0.785398$ is already close to the limit for $n\rightarrow\infty$.