Definite Integral $\int{x^2+1 \over x^4+1}$

Question

Evaluate $$\int_0^{\infty}{x^2+1 \over x^4+1}$$

I tried using Integration by parts , $$\frac{{x^3 \over 3 }+x}{x^4+1}+\int\frac{{x^3 \over 3 }+x}{(x^4+1)^2}.4x^3.dx$$

First term is zero

But it got me no where. Any hints.

Answer 1

• Divide the numerator by $x^2$ and you have $1+\frac{1}{x^2}$

• Divide the denominator by $x^2$ and you have $x^2+\frac{1}{x^2} =\bigl(x-\frac{1}{x}\bigr)^{2}+2$

Answer 2

Hint: $$\frac{x^2+1}{x^4+1}=\frac1{2\left(x^2+\sqrt2x+1\right)}+\frac1{2 \left(x^2-\sqrt2x+1\right)}.$$

Answer 3

Another approach: Separate out the integral into

$$ I = \int_0^\infty \frac{x^2}{x^4 + 1}\:dx + \int_0^\infty \frac{1}{x^4 + 1}\:dx $$

Both take the form of known integral

$$ \int_0^\infty \frac{x^k}{\left(x^n + a\right)^m}\:dx = a^{\frac{k +1}{n} -m}\frac{\Gamma\left(m - \frac{k + 1}{n}\right)\Gamma\left(\frac{k + 1}{n}\right)}{n\Gamma(m)} $$ if you wish to re-position your integrals into this form, take the following sequences of actions

(1) Substitute $x \mapsto x^4$

Then

(2) Substitute $x \mapsto \frac{1}{1 + x}$

This will put the integral into the form of the Beta Function

(3) Use the relationship between the Beta and Gamma Function: $$ B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} $$

And you're done.

For your specific case this then yields

\begin{align*} I &= \int_0^\infty \frac{x^2}{x^4 + 1}\:dx + \int_0^\infty \frac{1}{x^4 + 1}\:dx \\ &= 1^{\frac{2 +1}{4} -1}\frac{\Gamma\left(1 - \frac{2 + 1}{4}\right)\Gamma\left(\frac{2 + 1}{4}\right)}{4\Gamma(1)} + 1^{\frac{0 +1}{4} -1}\frac{\Gamma\left(1 - \frac{0 + 1}{4}\right)\Gamma\left(\frac{0 + 1}{4}\right)}{4\Gamma(1)} \\ &= \frac{\Gamma\left(1 - \frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)}{4} + \frac{\Gamma\left(1 - \frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{4} \end{align*}

Here we employ Euler's Reflection Formula \begin{equation*} \Gamma(z)\Gamma(1 - z) = \frac{\pi}{\sin\left(\pi z\right)} \qquad z \not \in \mathbb{Z} \end{equation*} Here we have $z = \frac{3}{4}, \frac{1}{4}$ which are both not integers and thus we can use the identity:

\begin{align*} I &= \frac{\Gamma\left(1 - \frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)}{4} + \frac{\Gamma\left(1 - \frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{4} = \frac{1}{4} \cdot \frac{\pi}{\sin\left(\pi \cdot \frac{3}{4} \right)} + \frac{1}{4} \cdot \frac{\pi}{\sin\left(\pi \cdot \frac{1}{4} \right)} \\ &= \frac{1}{4} \cdot \frac{\pi}{\frac{1}{\sqrt{2}}} + \frac{1}{4} \cdot \frac{\pi}{\frac{1}{\sqrt{2}}} = \frac{\pi}{\sqrt{2}} \end{align*}