Complete question: If $I_n$=$\int\limits_0^\infty\frac{x^{2n-2}}{(x^4-x^2+1)^n}dx$ and $I_5$=$\frac{p\pi}{q}$ where p and q are coprime natural numbers, then the value of 8p-q is what?
I have no clue how to go about this question. But I am thinking maybe $I_1$, $I_2$, $I_3$… are a part of some sequence and evauating a simpler integral will lead us to $I_5$ without actually solving for it. I got this idea since solving $I_1$ and $I_2$ is quite easy.
My level- High School.
Best Answer
Utilize $J(a)=\int_0^\infty\frac{1}{x^4-ax^2+1}dx=\frac\pi{2\sqrt{2-a}} $ to evaluate \begin{align} I_n=&\int_0^\infty\frac{x^{2n-2}}{(x^4-x^2+1)^n}dx = \frac1{(n-1)!}\frac{d^{n-1}J(a)}{da^{{n-1}}}\bigg|_{a=1} \end{align}
In particular \begin{align} I_5=&\frac1{4!}\frac{d^{4}J(a)}{da^{4}}\bigg|_{a=1} = \frac1{4!}\frac{d^{4}}{da^{{4}}} \bigg(\frac\pi{2\sqrt{2-a}}\bigg)\bigg|_{a=1}=\frac{35\pi}{256} \end{align} Thus, $8p-q=8(35)-256= 24.$