Let $a$ be a non-zero real number. Is it true that the value of $$\int\limits_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$$ is independent on $a$?
[Math] Is the integral $\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$ equal for all $a \neq 0$
calculusdefinite integralsintegration
Related Solutions
I believe this can also be solved using double integrals.
It is possible (if I remember correctly) to justify switching the order of integration to give the equality:
$$\int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dy \Bigg)\, dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$ Notice that $$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$
This leads us to
$$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$ Now the right hand side can be found easily, using integration by parts.
$$\begin{align*} I &= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} - y \int e^{-xy} \cos x \, dx\\ &= -e^{-xy}{\cos x} - y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big)\\ &= \frac{-ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2}. \end{align*}$$ Thus $$\int_{0}^{\infty} e^{-xy} \sin x \,dx = \frac{1}{1+y^2}$$ Thus $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$
This is not a complete solution.
First, we use a simple integral (we can get it by differentiating $x^b$ w.r.t. $b$ two times):
$$\int_0^1 x^b \ln^2 x ~d x=\frac{2}{(b+1)^3}$$
Using this integral, we also get:
$$\int_0^1 x^b \ln^2 x ~dx=\int^{\infty}_1 t^{-b-2} \ln^2 t~ dt=\frac{2}{(b+1)^3}$$
We separate the original integral into two parts:
$$ I=-\int\limits_0^1 \frac{\log^2 x \cos (ax)}{1-x^n} ~ dx+\int\limits_1^\infty \frac{\log^2 x \cos (ax)}{x^n-1} ~ dx=I_1+I_2 $$
Let's deal with the first one by using the modified integral from above:
$$\int_0^1 x^{pn} (a x)^{2q} \ln^2 x ~d x=\frac{2a^{2q}}{(pn+2q+1)^3}$$
Now we immediately get (using Taylor expansions for $\cos (ax)$ and $\frac{1}{1-x^n}$):
$$I_1=-2 \sum_{q=0}^{\infty} \frac{(-1)^q a^{2q}}{(2q)!} \sum_{p=0}^{\infty} \frac{1}{(pn+2q+1)^3}=\frac{1}{n^3} \sum_{q=0}^{\infty} \frac{(-1)^q a^{2q}}{(2q)!} \psi^{(2)} \left( \frac{2q+1}{n} \right)$$
Here $\psi^{(2)}$ is the second derivative of the digamma function.
This is a correct series solution for $I_1$, as we can see by comparing the two sides of the equation. See the plot $I_1(n)$ for $a=3$ below:
However, we can't use this method for $I_2$.
Let's see why (now we expand $\frac{1}{1-1/x^n}$):
$$\int^{\infty}_1 x^{-pn-n} (a x)^{2q} \ln^2 x~ dx=\frac{2a^{2q}}{(pn+n-2q-1)^3}$$
And now we see the problem. The denominator can become $0$ for infinitely many combinations of $p,q,n$, thus our integral would diverge badly.
Let's however, solve the case $a=0$, solution for which was provided in the OP ($n>1$).
$$I_2(a=0)=2 \sum_{p=0}^{\infty} \frac{1}{(pn+n-1)^3}=-\frac{1}{n^3} \psi^{(2)} \left(1- \frac{1}{n} \right)$$
Comparing with $I_1(a=0)$ and using the reflection formula for $\psi^{(2)}$:
$$\psi^{(2)}(1-x)-\psi^{(2)}(x)=2 \pi^3 \cot \left( \pi x \right) \csc^2 \left( \pi x \right)$$
we obtain:
$$I(a=0)=-\frac{2 \pi^3}{n^3} \cot \left( \pi \frac{2q+1}{n} \right) \csc^2 \left( \pi \frac{2q+1}{n} \right)$$
So, I'm not sure how to get the series solution for $I_2$, but I hope my answer helps you in some way.
Best Answer
Let $\mathcal{I}(a)$ denote the integral. Then $$ \begin{eqnarray} \mathcal{I}(a) &=& \int_0^1 \frac{\mathrm{d} x}{(1+x^2)(1+x^a)} + \int_1^\infty \frac{\mathrm{d} y}{(1+y^2)(1+y^a)} \\ &\stackrel{y=1/x}{=}& \int_0^1 \frac{\mathrm{d} x}{(1+x^2)(1+x^a)} + \int_0^1 \frac{x^a \mathrm{d} x}{(1+x^2)(1+x^a)} \\ &=& \int_0^1 \frac{1+x^a}{(1+x^2)(1+x^a)} \mathrm{d} x = \int_0^1 \frac{1}{1+x^2} \mathrm{d} x = \frac{\pi}{4} \end{eqnarray} $$
Thus $\mathcal{I}(a) = \frac{\pi}{4}$ for all $a$. I do not see a need to require $a$ to be non-zero.