# Approximate $\int_{0}^{\infty} \frac{\text{d} x}{1 + x^4}$

Now, I have been given this integral. And need to approximate it. My first idea was to use a Taylor series, but this series explodes, as x reaches infinity.

Does anyone know how to approximate improper integrals, (and this one in particular)?

I know I can use contour-integration to evaluate it exact, but I want to estimate it. Someone mentioned something about a Taylor expansion at infinity, but alas I have not learned about this.

$$\int_{0}^{\infty} \frac{\text{d} x}{1 + x^4}$$

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We can compute this integral without contour: math.stackexchange.com/questions/43457/… . –  Davide Giraudo Jan 15 '12 at 13:31
This integral can be computed exactly. More generally, $$\int_{0}^{\infty}\frac{1}{1+x^{n}}dx=\frac{1}{n}B( \frac{1}{n},1-\frac{1}{n} )=\frac{\pi}{n\sin\frac{\pi}{n}}$$ by making the change of variable $x^n=t$, where $B$ is the Beta function. See wikipedia. For $n=4$ you get the exact answer, $\dfrac{\pi\sqrt{2}}{4}$. –  digital-Ink Jan 15 '12 at 14:19
A Taylor expansion at infinity means that you'll expand in powers of $(\frac1x)$ (for $x$ large). Here $\frac{1}{1+x^4}=\frac{1}{x^4(1+1/x^4)}=\frac{1}{x^4}-\frac{1}{x^8}+O(\frac{1}{x‌​^{10}})$. –  Raymond Manzoni Jan 15 '12 at 14:23
Geesh... I already know how to evaluate this integral, I merely wanted to know how to approximate impropper integrals. I thought 1/x^4+1 would be a good example. Here is my stab at evaluating the integral exact i.imgur.com/PcXhd.gif. But this question was about approximations, not beta functions =) –  N3buchadnezzar Jan 15 '12 at 14:59

One way is to split integration range at $x=1$ and use geometric series approximation: $$\begin{eqnarray} \int_0^\infty \frac{\mathrm{d} x}{1+x^4} &=& \int_0^1 \frac{\mathrm{d} x}{1+x^4} +\int_1^\infty \frac{\mathrm{d} x}{1+x^4} \stackrel{x -> 1/x \text{ in second}}{=} \\ &=& \int_0^1 \frac{\mathrm{d} x}{1+x^4} +\int_0^1 \frac{x^2 \mathrm{d} x}{1+x^4} = \int_0^1 \frac{1+x^2}{1+x^4}\mathrm{d} x \\ &=& \int_0^1 \left( 1+ x^2-x^4 - x^6+x^8 + \cdots\right) \mathrm{d}x \\ &=& 1 + \frac{1}{3}-\frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \cdots = \frac{347}{315} + \cdots \end{eqnarray}$$ Now $347/315$ is approximately $1.10159$, while the exact answer is $\frac{\pi}{2\sqrt{2}} \approx 1.11072$.

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Transform your integral to have bounded endpoints. Here is one possible way $$\int{dx\over 1 + x^4} = \int_0^{\pi/2}{\sec^2(x)\, dx\over 1 + \tan^4(x)} = \int_0^{\pi/2} {\cos^2(x)\, dx\over \cos^4(x) + \sin^4(x)}.$$ This integral has a bounded integrand on a bounded interval. There are a panoply of ways to do this; choosing the cleverest one is an art.

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I can also transform it to $\int_{0}^{1} \frac{1+t^2}{1+t^4} \text{d}x$. Thanks for the suggestion about using substitutions =) will accept this in a few days, if nothing else revolutionary "cleverest" transformations shows up. –  N3buchadnezzar Jan 15 '12 at 15:01

Well, $1+x^4 > x^4$ for large $x$ so $$\int_a^{\infty}\frac{1}{1+x^4}dx \leq \int_a^{\infty}\frac{1}{x^4}dx =\frac{1}{3a^3}.$$ Now choose $a$ to be large enough to get this portion of the integral small (say less than $\varepsilon/2$). The remaining integral from $0$ to $a$ can be dealt with using standard numerical techniques.

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@Dylan Thanks for the observation. –  Mark McClure Jan 15 '12 at 14:00