Integrate $\frac{1}{(1+x^2)(1+x^c)}$ from $0$ to $\infty$ for any $c$.

Question

The question is to evaluate $$ \int_0^\infty \frac{dx}{(1+x^2)(1+x^c)} $$ for arbitrary $c\geq0$. Here are my attempts:

(The methods behave somewhat differently for $c=0$ but that case is trivial so I will assume $c>0$)

Integrate by parts: \begin{align*} \int_0^\infty \frac{dx}{(1+x^2)(1+x^c)}&=\arctan(x)\frac{1}{1+x^c}\Big|_0^\infty+\int_0^\infty\arctan(x)\frac{cx^{c-1}}{(1+x^c)^2}dx\\ &=0+\int_0^\infty\arctan(x)\frac{cx^{c-1}}{(1+x^c)^2}dx\\ &=\int_0^\infty\frac{\arctan(x^{1/c})}{(1+x)^2}dx. \end{align*} This looks just as bad, if not worse, as before...

Partial fractions: good for small values of $c$, but seems impossible as $c$ gets large.

Along the same line as partial fractions, but I tried the Ostragodski method, which only helped simplify things for $c=2$.

Try and show it doesn't depend on $c$:

For $c=1$, $c=2$, and $c=3$, the integral evaluates to $\pi/4$. My gut tells me this isn't a coincidence, so define $$ f(c)=\int_0^\infty\frac{dx}{(1+x^2)(1+x^c)} $$ so that $$ f'(c)=\int_0^\infty\frac{-x^c\log(x)}{(1+x^2)(1+x^c)^2}dx. $$ It would be great if this evaluated to $0$ for arbitrary $c$, but I don't see a nice way to show that it does.

Any hints would be greatly appreciated. Thanks in advance.

Answer 1

For $c\ge0$,

$$\begin{align} \mathcal{I}{\left(c\right)} &=\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(1+x^{2}\right)\left(1+x^{c}\right)}\\ &=\int_{0}^{1}\frac{\mathrm{d}x}{\left(1+x^{2}\right)\left(1+x^{c}\right)}+\int_{1}^{\infty}\frac{\mathrm{d}x}{\left(1+x^{2}\right)\left(1+x^{c}\right)}\\ &=\int_{0}^{1}\frac{\mathrm{d}x}{\left(1+x^{2}\right)\left(1+x^{c}\right)}+\int_{0}^{1}\frac{y^{c}\,\mathrm{d}y}{\left(1+y^{2}\right)\left(1+y^{c}\right)};~~~\small{\left[\frac{1}{x}=y\right]}\\ &=\int_{0}^{1}\frac{\left(1+x^{c}\right)\,\mathrm{d}x}{\left(1+x^{2}\right)\left(1+x^{c}\right)}\\ &=\int_{0}^{1}\frac{\mathrm{d}x}{1+x^2}\\ &=\frac{\pi}{4}.\\ \end{align}$$

Answer 2

Sub $x=\tan{t}$. Then the integral is

$$\int_0^{\pi/2} \frac{dt}{1+\tan^c{t}} $$

Then see this answer.

ADDENDUM

For those too tired to click the link:

Use the fact that

$$\tan{\left (\frac{\pi}{2}-x\right)} = \frac{1}{\tan{x}}$$

i.e.,

$$\frac1{1+\tan^{c}{x}} = 1-\frac{\tan^{c}{x}}{1+\tan^{c}{x}} = 1-\frac1{1+\frac1{\tan^{c}{x}}} = 1-\frac1{1+\tan^{c}{\left (\frac{\pi}{2}-x\right)}}$$

Therefore, if the sought-after integral is $I$, then

$$I = \frac{\pi}{2}-I$$

and...

ADDENDUM II

It should be noted that the above result is valid for any real value of $c$, not just positive values.

Answer 3

$$I(0)=\frac{\pi}{4}$$

$$\frac{d I}{dc}=-\int_0^{\infty}\frac{x^c\log x}{(1+x^2)(1+x^c)^2}dx$$

$$\frac{d I}{dc}\stackrel{x\rightarrow 1/x}{=}+\int_0^{\infty}\frac{x^c\log x}{(1+x^2)(1+x^c)^2}dx =-\frac{d I}{dc}$$

$$\frac{d I}{dc}=0 \Rightarrow I(c)=const=\frac{\pi}{4}$$

Answer 4

$$ I=\int_0^\infty \frac{dx}{(1+x^2)(1+x^c)} ={{\pi} \over 4}$$ For any $c \gt 0$

Substitute $$x=\tan(u)$$

Cancel the $\sec(x)^2$

We'll get $$\int_0^{\pi/2} {{1} \over {1+\tan^c(u)}} \ du$$

Symmetry can evaluate this integral $$tan(\pi/2-u)=cot(u)$$

Change variables to $v=\pi/2-u$

We'll get $$I=\int_0^{\pi/2} {{\tan^c(u)} \over {tan^c(u)+1}} \ du$$ Add this to the first change of variables. Note the terms cancel and we get. $$2 \cdot I=\int_0^{\pi/2} du$$ Which is $2 \cdot I=\pi/2$, which is the integral times two. Thus,

$$ \int_0^\infty \frac{dx}{(1+x^2)(1+x^c)} ={{\pi} \over 4}$$ For any $c$ actually

This trick can is further discussed here in the second answer.

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