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. ${}\qquad{}$ Commented Sep 18, 2015 at 6:05
• Hi junatheron I got $$\frac{\pi } {2}\frac{1} {{\sqrt {3 + \sqrt 2 } }}$$ can you verify my answer if the answer is wrong then I don't want to upload it Commented Nov 20, 2023 at 13:36

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}

We'll use a standard contour integration argument, and we'll change the parameter in a way convenient for our computation.

By symmetry, $$I = \frac14 \int_{-\infty}^\infty \frac{(1 + x^2) \,dx}{1 + a x^2 + x^4}.$$ For $$a > 2$$, we can write $$a = 2 \cosh 2 t$$ for some $$t > 0$$; we denote the integrand by $$f_t := \frac{1 + x^2}{1 + 2 \cosh 2 t \cdot x^2 + x^4}$$

The poles of $$f_t$$ in the upper half-plane are at $$i e^{\pm t}$$, and applying the Residue Theorem to the boundary of a half-disk of radius $$R$$ in the upper half-plane centered at the origin gives that \begin{align*} 4 I = \int_{-\infty}^\infty f_t \,dx &= 2\pi i [\operatorname{Res}(f_t, i e^t) + \operatorname{Res}(f_t, i e^{-t})] \\ &= 2\pi i \left(-\frac{i}4 \operatorname{sech} t -\frac{i}4 \operatorname{sech} t\right) \\ &= \pi \operatorname{sech} t \\ &= \frac{2\pi}{\sqrt{2 + a}} . \end{align*} So, $$\boxed{I = \frac{\pi}{2 \sqrt{2 + a}}} .$$

\begin{align} I &= \int\limits_0^{+\infty} \frac{dx}{(x^4 + (1 + \sqrt{2})x^2 + 1)(x^{100} - x^{98} + \ldots + 1)} \\ &= \frac{\pi}{2\sqrt{3 + \sqrt{2}}} \\ I &= \int\limits_0^{+\infty} \frac{dx}{(x^4 + (1 + \sqrt{2})x^2 + 1)(x^{100} - x^{98} + \ldots + 1)} \underbrace{=}_{x = \frac{1}{y}} \int\limits_0^{+\infty} \frac{y^{102}}{(y^4 + (1 + \sqrt{2})y^2 + 1)(y^{100} - y^{98} + \ldots + 1)} \,dy \\ I &= \frac{1}{2} \int\limits_0^{+\infty} \left(\frac{1}{(x^4 + (1 + \sqrt{2})x^2 + 1)(x^{100} - x^{98} + \ldots + 1)} + \frac{x^{102}}{(x^4 + (1 + \sqrt{2})x^2 + 1)(x^{100} - x^{98} + \ldots + 1)}\right) \,dx \\ &= \frac{1}{2} \int\limits_0^{+\infty} \frac{1 + x^{102}}{(x^4 + (1 + \sqrt{2})x^2 + 1)(x^{100} - x^{98} + \ldots + 1)} \,dx \\ x^{100} - x^{98} + \ldots + 1 &= \frac{1 + x^{2 \cdot 51}}{1 + x^2} = \frac{1 + x^{102}}{1 + x^2} \\ &= \frac{1}{2}\int\limits_0^{+\infty} \frac{1 + x^2}{x^4 + (1 + \sqrt{2})x^2 + 1} \,dx = \frac{1}{2}\int\limits_0^{+\infty} \frac{1 + x^2}{x^2(x^2 + \sqrt{2}) + x^2 + 1} \,dx \\ \underbrace{=}_{x^2 + 1 = u} \frac{1}{2}\int\limits_1^{+\infty} \frac{u}{(u - 1)(u - 1 + \sqrt{2}) + u}\frac{du}{2\sqrt{u - 1}} \underbrace{=}_{u - 1 = y} \frac{1}{4}\int\limits_0^{+\infty} \frac{y + 1}{y(y + \sqrt{2}) + y + 1}\frac{du}{\sqrt{y}} \\ &= \frac{1}{4}\int\limits_0^{+\infty} \frac{y + 1}{y^2 + y(1 + \sqrt{2}) + 1}\frac{dy}{\sqrt{y}} \underbrace{=}_{y = \tan^2 \theta} \frac{1}{2}\int\limits_0^{\frac{\pi}{2}} \frac{\tan^2 \theta + 1}{\tan^4 \theta + \tan^2 \theta(1 + \sqrt{2}) + 1}\frac{\tan \theta \sec^2 \theta d\theta}{\sqrt{\tan^2 \theta}} \\ &= \frac{1}{2}\int\limits_0^{\frac{\pi}{2}} \frac{\sec^4 \theta}{\sec^2 \theta + \tan^4 \theta\left(1 + \frac{\sqrt{2}}{\tan^2 \theta}\right)}d\theta = \frac{1}{2}\int\limits_0^{\frac{\pi}{2}} \frac{d\theta}{\cos^2 \theta + \sin^4 \theta + \sqrt{2}\cos^2 \theta \sin^2 \theta} \\ &= \frac{1}{2}\int\limits_0^{\frac{\pi}{2}} \frac{d\theta}{1 + (\sqrt{2} - 1)\sin^2 \theta + (1 - \sqrt{2})\sin^4 \theta} = \frac{1}{{2(1 - \sqrt{2})}}\int\limits_0^{\frac{\pi}{2}} \frac{d\theta}{-(1 + \sqrt{2}) - \sin^2 \theta(1 - \sin^2 \theta)} \\ &= 2(\sqrt{2} + 1)\int\limits_0^{\frac{\pi}{2}} \frac{d\theta}{4(\sqrt{2} + 1) + \sin^2 2\theta} = \frac{1}{2}\int\limits_0^{\frac{\pi}{2}} \frac{d\theta}{1 + \frac{\sin^2 2\theta}{4(\sqrt{2} + 1)}} \\ &= \frac{1}{2}\int\limits_0^{\frac{\pi}{2}} \sum\limits_{k=0}^{+\infty} (-1)^k \left(\frac{\sin^2 2\theta}{4(\sqrt{2} + 1)}\right)^k d\theta \\ &= \frac{1}{2}\sum\limits_{k=0}^{+\infty} \frac{(-1)^k}{2^{2k}(\sqrt{2} + 1)^k} \int\limits_0^{\frac{\pi}{2}} \sin^{2k} 2\theta d\theta \\ &= \frac{1}{2}\sum\limits_{k=0}^{+\infty} \frac{(-1)^k}{(\sqrt{2} + 1)^k} \int\limits_0^{\frac{\pi}{2}} \sin^{2k} \theta \cos^{2k} \theta d\theta \\ &= \frac{1}{4}\sum\limits_{k=0}^{+\infty} \frac{(-1)^k}{(\sqrt{2} + 1)^k} 2\int\limits_0^{\frac{\pi}{2}} \sin^{2\left(k + \frac{1}{2}\right) - 1} \theta \cos^{2\left(k + \frac{1}{2}\right) - 1} \theta d\theta \\ &= \frac{1}{4}\sum\limits_{k=0}^{+\infty} \frac{(-1)^k}{(\sqrt{2} + 1)^k} \beta\left(k + \frac{1}{2}, k + \frac{1}{2}\right) \end{align}

$$\displaystyle{ = \frac{1} {4}\sum\limits_{k = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{\left( {\sqrt 2 + 1} \right)^k }}\frac{{\Gamma ^2 \left( {k + \frac{1} {2}} \right)}} {{\Gamma \left( {2k + 1} \right)}}} = \frac{1} {4}\sum\limits_{k = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{\left( {\sqrt 2 + 1} \right)^k }}\frac{{\left( {\frac{{\sqrt \pi }} {{2^{2k - 1} }}\frac{{\Gamma \left( {2k} \right)}} {{\Gamma \left( k \right)}}} \right)^2 }} {{\Gamma \left( {2k + 1} \right)}}} }$$ $$\displaystyle{ = \frac{1} {4}\sum\limits_{k = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{\left( {\sqrt 2 + 1} \right)^k }}\frac{{\Gamma ^2 \left( {k + \frac{1} {2}} \right)}} {{\Gamma \left( {2k + 1} \right)}}} = \pi \sum\limits_{k = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{4^{2k} \left( {\sqrt 2 + 1} \right)^k }}\frac{{\Gamma ^2 \left( {2k} \right)}} {{\Gamma ^2 \left( k \right)\Gamma \left( {2k + 1} \right)}}} }$$ $$\displaystyle{ = \frac{\pi } {4}\sum\limits_{k = 0}^{ + \infty } {\left( { - \frac{1} {{4^2 \left( {\sqrt 2 + 1} \right)}}} \right)^k \frac{{\left( {2k} \right)!}} {{\left( {k!} \right)^2 }}} = \frac{\pi } {4}\frac{1} {{\sqrt {1 - 4 \cdot - \frac{1} {{16\left( {\sqrt 2 + 1} \right)}}} }} = \frac{\pi } {4}\frac{1} {{\sqrt {1 + \frac{{\sqrt 2 - 1}} {4}} }} = \frac{\pi } {2}\frac{1} {{\sqrt {3 + \sqrt 2 } }} }$$