# Integral $\int_0^\infty \frac{\log^2 x \cos ax}{x^n-1}dx$

Hi I am trying to calculate $$I:=\int\limits_0^\infty \frac{\log^2 x \cos (ax)}{x^n-1}\mathrm dx,\quad \Re(n)>1, \, a\in \mathbb{R}.$$ Note if we set $a=0$ we get a similar integral given by $$\int\limits_0^\infty \frac{\log^2 x }{x^n-1}\mathrm dx=-\frac{2\pi^3\cot(\pi/n)\csc^2(\pi/n)}{n^3},\quad \Re(n)>1.$$ I was trying to write I as $$I=\Re \bigg[\int\limits_0^\infty \frac{ e^{i ax}\log^2 x}{x^n-1}\mathrm dx\bigg]=\Re\bigg[\int\limits_\infty^0\frac{e^{iax}\log^2 x}{1-x^n}\mathrm dx\bigg]=\Re\bigg[\int\limits_\infty^0e^{iax}\log^2 x\sum_{m=0}^\infty x^{nm} \mathrm dx\bigg].$$ But was unsure of where to go from here. How can we calculate $I$? It is clear that this method is not going to work.

• Where did you find this integral? – Mhenni Benghorbal Apr 24 '14 at 4:21
• @MhenniBenghorbal I didn't "find" it anywhere, I am not sure what that even means though? – Jeff Faraci Apr 24 '14 at 4:24
• I mean where it came from? Or you are trying to make a generalization. – Mhenni Benghorbal Apr 24 '14 at 4:26
• @MhenniBenghorbal Oh, I am just trying to make a generalization my friend! I have not seen this in a paper, but I have seen these kinds of integrals growing up all the time – Jeff Faraci Apr 24 '14 at 4:37
• @sidht Consider the well known Gaussian integral $$\int_{-\infty}^\infty e^{-\alpha x^2 +\beta x}dx=\frac{\sqrt \pi}{\sqrt \alpha} e^{\beta^2/4\alpha},\quad \Re(\alpha)>0,$$ another integral with two parameters! This one is quite famous for obvious reason. It is good to get used to integrals with parameters – Jeff Faraci Apr 24 '14 at 16:39

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.

This is my failed attempt at this problem. Essentially too long for a comment and contains multiple mistakes.

$$\int_0^\infty \cos ax^2\text{ d}x=\sqrt\frac{\pi}{8a}$$

$$\frac{\partial^{-4mn}}{\partial a^{-4mn}}\int_0^\infty {\cos ax^2}=\int_0^\infty \frac{\cos ax^2}{x^{8mn}}\text{ d}x=\frac{\pi}{\sqrt 8}\frac{a^{4mn-1/2}}{\Gamma (4mn+1/2)}$$

Which is true because of this question: Derivative of a negative order?

$$\sum_{m=1}^\infty \int_0^\infty \frac{\partial^{-4mn}}{\partial a^{-4mn}}\cos ax^2 \, dx=\int_0^\infty \left(\frac{\cos ax^2}{x^{8n}}+\frac{\cos ax^2}{x^{16n}}+\frac{\cos ax^2}{x^{24n}}\cdots\right)\text{ d}x=\int_0^\infty \frac{\cos ax^2}{x^{8n}-1}\text{ d}x$$

Now, substitute $x\to\sqrt x, \text{ d}x\to\frac{\text{ d}x}{2\sqrt x}$ and assume that $n=\frac{1}{4}k$

And we get: $$\frac{1}{2}\int_0^\infty \frac {\cos ax}{\sqrt x (x^k-1)} \text{ d}x=\frac{\pi}{\sqrt 8}\sum_{m=1}^\infty \frac{a^{mn-1/2}}{\Gamma(mk+1/2)}$$

Maybe there is a part of this proof that you can deviate from.