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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?

HASNAIN
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  • The "clever" way to do it is to use the residue theorem. – Ian Sep 07 '17 at 04:04
  • Why do you think it is related to the Fourier transform ? You can compute $\displaystyle\int_0^x \frac{dt}{t^4+1}$ using partial fraction decomposition : $\displaystyle\frac{1}{t^4+1} =\frac{1}{\prod_{k=1}^4 (t-e^{(2k+1)i\pi/4})}= \sum_{k=1}^4 \frac{a_k}{t-e^{(2k+1)i \pi /4}}$ where $\frac{1}{a_k} = f'(e^{(2k+1)i\pi/4}),f(t) = t^4+1$ – reuns Sep 07 '17 at 04:13
  • @reuns i have done many integration with the help of fourier transform properties.sorry if wrongly tagged it – HASNAIN Sep 07 '17 at 04:25
  • @Ian can you please explain it how can i use residue theorem ? – HASNAIN Sep 07 '17 at 04:26
  • This problem, after already changing variables to $w/B$, appears in basically any complex analysis book as an exercise with the residue theorem. It is quite standard, but you do need to know some background to make sense of the technique. – Ian Sep 07 '17 at 04:28
  • In this particular case, the residue theorem is just another name for partial fraction decomposition. @Ian – reuns Sep 07 '17 at 04:36
  • @reuns Not really. Partial fractions gives 4 terms but they do not all actually contribute with these bounds. Plus you have to make sure to keep your complex logarithms straight. – Ian Sep 07 '17 at 04:44
  • @Ian The residue theorem = partial fraction decomposition + Cauchy integral theorem. – reuns Sep 07 '17 at 04:46
  • Why don't you both give it as an answer? Please. – HASNAIN Sep 07 '17 at 04:48
  • @USER786 See my 1st comment and Partial_fraction_decomposition – reuns Sep 07 '17 at 05:12
  • @reuns The reason I disagree is that if I think of "partial fraction decomposition" then I assume that you will calculate all the coefficients. But you only really need to calculate the coefficients corresponding to residues inside the relevant contour. So like here, $\frac{1}{x^4+1}=\sum_{j=0}^3 \frac{a_j}{x-e^{(2j+1)i\pi/4}}$, but you only need $a_0$ to evaluate the integral. – Ian Sep 07 '17 at 13:19

3 Answers3

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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.

Nosrati
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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*}

Tianlalu
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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 $I=\frac{\pi i}{2(1-i)} e^{-3 i \pi/4}=\frac{\pi}{2\sqrt{2}}$.

Ian
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