Integration of $\int_{0}^{\infty}\frac{dw}{1+\left ( \frac{w}{B} \right )^4}$ with the help of signal properties.

Question

Find the integral $$\int_{0}^{\infty}\frac{dw}{1+\left ( \frac{w}{B} \right )^4}$$ where $B$ is a constant.

This integration i tried by normal method that gives the result $\frac{\pi B}{2\sqrt2}$ but it goes so lengthy like this

is there any signal properties that i use to solve it in simple steps?

Answer 1

With substitution $x^4=t$ and again with $\dfrac{1}{1+t}=u$ we have \begin{align} I &= B\int_0^\infty\dfrac{1}{1+x^4}dx \\ &= \dfrac{B}{4}\int_0^\infty\dfrac{t^{-\frac34}}{1+t}dt \\ &= \dfrac{B}{4}\int_0^1 u^{-\frac14}(1-u)^{-\frac34}du\\ &= \dfrac{B}{4}\beta(\frac14,\frac34) \\ &= \dfrac{B}{4}\dfrac{\Gamma(\dfrac14)\Gamma(\dfrac34)}{\Gamma(1)} \\ &= \dfrac{B}{4}\dfrac{\pi}{\sin\frac14\pi} \\ &= \color{blue}{\dfrac{\pi B}{2\sqrt{2}}} \end{align} where $\beta(x,y)$ is beta function.

Answer 2

Just want to share a "smart" method.

(Too long for comment)

Here is a fast way to evaluate the indefinite integral.

Sub $ t=Bx$,

Then, \begin{align*} \int \frac 1{1+x^4}\,dx&=\frac 12\left(\int \frac {1+x^2}{1+x^4}\,dx+\int \frac {1-x^2}{1+x^4}\,dx\right)\\ &=\frac 12\left(\int \frac {\color{red}{\left(\frac 1 {x^2}+1\right)dx}}{\frac 1{x^2}+x^2}+\int \frac {\color{blue}{\left(\frac 1 {x^2}-1\right)dx}}{\frac 1{x^2}+x^2}\right)\\ &=\frac 12\left(\int \frac {\color{red}{d\left(x-\frac 1 x\right)}}{\left(x-\frac 1 x\right)^2+2}+\int \frac {\color{blue}{-d\left(x+\frac 1 x\right)}}{\left(x+\frac 1 x\right)^2-2}\right)\qquad\quad\\ &= \frac 12\left(\frac 1{\sqrt2} \tan^{-1}\left(\frac {x-\frac 1 x} {\sqrt2}\right)+\frac 1{\sqrt2} \tanh^{-1}\left(\frac {x+\frac 1 x} {\sqrt2}\right)\right). \end{align*}

Answer 3

A way to do the integral over the whole half-line with the residue theorem is the following. Consider the contour going straight from $0$ to $R$, following a circular arc from $R$ to $iR$, and then going straight from $iR$ to $0$. The desired integral $I$ is the integral on the first part of the contour. The last piece is

$$\int_R^0 \frac{1}{(ix)^4+1} d(ix)=i \int_R^0 \frac{1}{x^4+1} dx = -iI.$$

The middle piece can be shown using an "ML estimate" to go to zero as $R \to \infty$. Now the only pole of the integrand inside the contour is at $e^{i\pi/4}$, so the residue theorem tells you that the integral around the contour is

$$J=(1-i)I=2\pi i\operatorname{Res} \left ( \frac{1}{z^4+1},e^{i \pi/4} \right ).$$

This residue is $\lim_{z \to e^{i\pi/4}} \frac{z-e^{i\pi/4}}{z^4+1}=\frac{1}{4} e^{-3 i \pi/4}$, so $I=\frac{\pi i}{2(1-i)} e^{-3 i \pi/4}=\frac{\pi}{2\sqrt{2}}$.