Finding $\int_{0}^{\infty }\frac{1}{1+x^4}dx$

Question

finding $$\int_{0}^{\infty }\frac{1}{1+x^4}dx$$ My attempt is:

let $x=\sqrt{u}$

$dx=\frac{1}{2\sqrt{u}}$ $$\int_{0}^{\infty }\frac{1}{2\sqrt{u}(1+u^2)}du$$ here I stopped because I don't know how to complete this solution. Any help please.

Answer 1

One way to deal with this is to write $$ 1+x^4 = (1 + 2x^2+x^4) - 2x^2 = (1+x^2)^2 - (x\sqrt 2)^2 = (1+x^2+x\sqrt2)(1+x^2-x\sqrt 2) $$ Then use partial fractions: $$ \frac {Ax+B} {1+x^2+x\sqrt2} + \frac {Cx+D} {1+x^2-x\sqrt2} $$ Notice that you can tell that these quadratic polynomials are irreducible because their discriminants $b^2-4ac = (\sqrt 2)^2 - 4\cdot1\cdot1$ are negative. You have to do a bit of algebra to find $A$, $B$, $C$, and $D$.

If you let $u=x^2+x\sqrt2+1$ then $du=(2x+\sqrt2)\,dx$ and $$ \frac{Ax+B}{x^2+x\sqrt2+1} \text{ becomes }\frac{\frac A 2 (2x+\sqrt 2)}{x^2+x\sqrt2+1} + \frac{\text{some consant}}{x^2+x\sqrt2+1} $$ The substitution above handles the first fraction. The second one is done by the substitution below: \begin{align} \int \frac {dx} {1+x^2+x\sqrt2} & = \int \frac{dx}{\left(x^2+x\sqrt2+\frac 1 2 \right) + \frac 1 2} \qquad(\text{completing the square}) \\[8pt] & = \int \frac {dx}{\left(x + \frac 1 {\sqrt2}\right)^2 + \frac 1 2} = \int \frac{2\,dx}{\left(x\sqrt2 + 1\right)^2 + 1}. \end{align} Then use the substitution \begin{align} x\sqrt2+1 & = \tan\theta \\[6pt] \text{so that }\left(x\sqrt2+1\right)^2 + 1 & = \sec^2\theta \\[6pt] \text{and }\sqrt 2\,dx & = \sec^2\theta\,d\theta \end{align} and the integral above becomes $$ \int \sqrt2\,d\theta = \sqrt 2\ \theta = \sqrt 2\ \arctan(x\sqrt2 + 1) +C $$ etc.

Perhaps I should add that the first time I ever saw this integral I did it by a more straightforward method and consequently saw how to do what I did above, but some people don't like that more straightforward method because it involves complex numbers. That method is this: to factor $x^4+1$ we solve $x^4+1=0$, getting \begin{align} x^4 & = -1 \\[6pt] x^2 & = \pm i \end{align} To find square roots of $i$, notice that $i = \cos90^\circ+i\sin90^\circ$, so its square roots must be $\pm(\cos45^\circ+i\sin45^\circ)= \pm\left(\frac{1+i}{\sqrt2}\right)$ and similarly square roots of $-i$ are $\pm(\cos45^\circ-i\sin45^\circ)= \pm\left(\frac{1-i}{\sqrt2}\right)$ Now we have $$ x^4+1=\underbrace{\left(x-\frac{1+i}{\sqrt2}\right)\left(x-\frac{1-i}{\sqrt2}\right)}_\text{conjugates}\ \underbrace{\left(x-\frac{-1+i}{\sqrt2}\right)\left(x-\frac{-1-i}{\sqrt2}\right)}_\text{conjugates}. $$ When conjugates are multiplied, the imaginary parts cancel out: $$ x^4+1 = (x^2-x\sqrt2+1)(x^2+x\sqrt2 + 1). $$

Answer 2

$$\int_0^\infty\frac{dx}{1+x^4}=\Re\left({\int_0^\infty\frac1{1-ix^2}\,dx}\right).$$

Then by decomposition , $$\int_0^\infty\frac{dx}{1-ix^2}=\frac12\int_0^\infty\left(\frac1{1-\sqrt ix}+\frac1{1+\sqrt ix}\right)\,dx=\frac1{2\sqrt i}\ln\left(\frac{1+\sqrt ix}{1-\sqrt ix}\right)\Big|_0^\infty=\frac{i\pi-0}{2\sqrt i},$$

$$\color{green}{I=\Re\left(\frac{\sqrt i\pi}2\right)=\frac\pi{\sqrt8}}.$$

Answer 3

$$\int_{0}^{\infty }\frac{1}{1+x^4}dx=$$ $$\int_{0}^{\infty }\left(\frac{\sqrt{2}x-2}{4(-x^2+\sqrt{2}x-1)}+\frac{\sqrt{2}x+2}{4(x^2+\sqrt{2}x+1)}\right)dx$$

The antiderivative of $\left(\frac{\sqrt{2}x-2}{4(-x^2+\sqrt{2}x-1)}+\frac{\sqrt{2}x+2}{4(x^2+\sqrt{2}x+1)}\right) $ is:

$$\frac{1}{4\sqrt{2}}(-\log(x^2-\sqrt{2}x+1)+\log(x^2+\sqrt{2}x+1)-2\tan^{-1}(1-\sqrt{2}x)+2\tan^{-1}(\sqrt{2}x+1))$$

So we got the limit of b goes to infinity:

$$\left[\frac{1}{4\sqrt{2}}(-\log(x^2-\sqrt{2}x+1)+\log(x^2+\sqrt{2}x+1)-2\tan^{-1}(1-\sqrt{2}x)+2\tan^{-1}(\sqrt{2}x+1))\right]^b_0$$

So we get:

$$\int_{0}^{\infty }\frac{1}{1+x^4}dx=\frac{\pi}{2\sqrt{2}}$$

Answer 4

Method 1:

Because we have an even integrand, we can rewrite

$$ I=\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{1+x^4} $$

Now we are ready to use residue theorem. A close inspection of the denominator shows that there are 2 poles in the upper half plane :

$$ x_{\pm}=\frac{\pm1+i}{\sqrt{2}} $$

Choosing a big semicircle in the uhp as an integration contour(the arc will vanish because the the integrand falls of as $1/R^3$ as $R\rightarrow \infty$ ) we get

$$ I= \frac{1}{2}\times2 \pi i \times[\text{Res}(x_+)+\text{Res}(x_-)]=\frac{\pi}{2\sqrt{2}} $$

Method 2:

Another (maybe too) sophisticated approach would be as follows

Write $x^4=q, dx= \frac{1}{4}q^{-3/4}dq$ and we get

$$ I=\frac{1}{4}\int_{0}^{\infty}\frac{1}{(1+x)x^{3/4}} $$

which yields

$$ I=\frac{1}{4}B(\frac{1}{4},\frac{3}{4}) $$

where we used one of the representations of Euler's Beta function. Using the fact the reflection formula for the Gamma function ($\Gamma(1-x)\Gamma(x)=\frac{\pi}{\sin(\pi x)}$) it follows that

$$ I= \frac{\pi}{\sin(3\pi/ 4)}= \frac{\pi}{2 \sqrt{2}} $$

Answer 5

Here's an approachable method without complex analysis.

Notice that you can write your integral as (which i leave to you why?)

$$\frac{1}{2}\int \frac{1+x^2}{1+x^4}\cdot dx+\frac{1}{2}\int \frac{1-x^2}{1+x^4}\cdot dx $$

I will only solve one of these and the other one for you.

Let me take

$$\int \frac{1-x^2}{1+x^4}\cdot dx$$

Taking $x$ common,

$$\int \frac{\frac{1}{x^2}-1}{\frac{1}{x^2}+x^2}\cdot dx $$

Now use the substitution $x-\frac{1}{x}=t$

Find $dt$ and also square the substitution to see what you can do for the denominator.

$$\left(1-\frac{1}{x^2}\right))\,dx=dt$$

and $x^2+\frac{1}{x^2}=t^2+2$

Gives the integral

$$\int \frac{-dt}{t^2+2}$$

This is an trivial integral of $\tan^{-1}$ And the other one i suspect is gonna be of the form $\frac{c}{x^2-a^2}$ probably a log but i leave that to you to try. :)