My brother’s friend gave me the following wicked integral with a beautiful result

\begin{equation}

{\Large\int_0^\infty} \frac{dx}{\sqrt{x} \bigg[x^2+\left(1+2\sqrt{2}\right)x+1\bigg] \bigg[1-x+x^2-x^3+\cdots+x^{50}\bigg]}={\large\left(\sqrt{2}-1\right)\pi}

\end{equation}He claimed the above integral can be generalised to the following form

\begin{equation}

{\Large\int_0^\infty} \frac{dx}{\sqrt{x} \bigg[x^2+ax+1\bigg] \bigg[1-x+x^2-x^3+\cdots+(-x)^{n}\bigg]}=\ldots

\end{equation}

This is a challenging problem. How to prove it and what is the closed-form of the general integral?

**Answer**

Indeed let

$$

I(n,a)=\int_0^\infty\frac{dx}{\sqrt{x}(1+ax+x^2)(\sum_{k=0}^n(-x)^k)}

$$

The change of variables $x\leftarrow 1/x$ yields

$$

I(n,a)=\int_0^\infty\frac{(-1)^nx^{n+1}dx}{ \sqrt{x}(1+ax+x^2)(\sum_{k=0}^n(-x)^k)}

$$

Thus

$$

2I(n,a)=\int_0^\infty\frac{1+x}{\sqrt{x}(1+ax+x^2)}dx=

2\int_0^\infty\frac{1+t^2}{ 1+at^2+t^4}dt

$$

Or equivalently, setting $u=t-1/t$,

$$

I(n,a)=

\int_{-\infty}^\infty\frac{du}{ 2+a+u^2} =\frac{\pi}{\sqrt{2+a}}.

$$

**Attribution***Source : Link , Question Author : Anastasiya-Romanova 秀 , Answer Author : Omran Kouba*