# Evaluation of $\int_0^\infty \frac{1}{\left[x^4+(1+\sqrt{2})x^2+1\right]\cdot \left[x^{100}-x^{98}+\cdots+1\right]}dx$

Evaluation of $\displaystyle \int_0^\infty \frac{1}{\left[x^4+(1+\sqrt{2})x^2+1\right]\cdot \left[x^{100}-x^{98}+\cdots+1\right]}dx$

$\bf{My\; Try::}$ Let $$I= \int_{0}^{\infty}\frac{1}{\left[x^4+(1+\sqrt{2})x^2+1\right]\cdot \left[x^{100}-x^{98}+\cdots+1\right]}dx \tag 1$$

Let $\displaystyle x=\frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}$ and changing limit, we get

So $$I = \int_0^\infty \frac{t^{102}}{\left[t^4+(1+\sqrt{2})t^2+1\right]\cdot \left[t^{100}-t^{98}+\cdots+1\right]}dt$$

So $$\displaystyle I = \int_0^\infty \frac{x^{102}}{\left[x^4+(1+\sqrt{2})t^2+1\right]\cdot \left[x^{100}-x^{98}+\cdots+1\right]}dx \tag 2$$

So we get $$2I = \int_0^\infty \frac{1+x^{102}}{\left[x^4+(1+\sqrt{2})x^2+1\right]\cdot \left[x^{100}-x^{98}+\cdots+1\right]} \,dx$$

Now Using Geometric Progression series,

We can write $$1-x^2+x^4-\cdots-x^{98}+x^{100} = \left(\frac{x^{102+1}}{1+x^2}\right)$$

so we get $$2I = \int_0^\infty \frac{1+x^2}{x^4+ax^2+1}dx\;,$$ Where $a=(\sqrt{2}+1)$

So we get $$2I = \int_0^\infty \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+\left(\sqrt{a+2}\right)^2} dx = \frac{1}{\sqrt{a+2}}\left[\tan^{-1}\left(\frac{x^2-1}{x\cdot \sqrt{a+2}}\right)\right]_0^\infty$$

So we get $$2I = \frac{\pi}{\sqrt{a+2}}\Rightarrow I = \frac{\pi}{2\sqrt{3+\sqrt{2}}}$$

My Question is can we solve the Integral $\bf{\displaystyle \int_0^\infty \frac{1+x^2}{x^4+ax^2+1}dx}$ Using any other Method

Means Using Complex analysis or any other.

Thanks.

• Why do you use \displaystyle in the context of an actual display? (And once you actually wrote \displaystyle twice in a row. ${}\qquad{}$ – Michael Hardy Sep 18 '15 at 6:05

## 1 Answer

I think your way of evaluating the integral is very short and elegant. Nevertheless, you can do partial fraction decomposition. You'll end up with (as long as I did not do any mistakes) $$\frac{1}{\sqrt{2+a}}\biggl[\arctan\Bigl(\frac{\sqrt{2-a}+2x}{\sqrt{2+a}}\Bigr)+\arctan\Bigl(\frac{\sqrt{2-a}-2x}{\sqrt{2+a}}\Bigr)\biggr]$$ as a primitive.

Edit That way of writing it is not so good, since $a>2$. It is better to write $$x^4+ax^2+1=\Bigl(x^2+\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\Bigr)\Bigl(x^2+\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\Bigr)$$ and then do partial fraction decomposition. The resulting primitive then reads (watch out for typos!) \begin{aligned} \frac{1}{\sqrt{2}\sqrt{a^2-4}}\biggl[&\frac{2-a+\sqrt{a^2-4}}{\sqrt{a-\sqrt{a^2-4}}}\arctan\Bigl(\frac{\sqrt{2}x}{\sqrt{a-\sqrt{a^2-4}}}\Bigr)\\ &\qquad+\frac{a-2+\sqrt{a^2-4}}{\sqrt{a+\sqrt{a^2-4}}}\arctan\Bigl(\frac{\sqrt{2}x}{\sqrt{a+\sqrt{a^2-4}}}\Bigr)\biggr]. \end{aligned}