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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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    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 http://i.imgur.com/PcXhd.gif. But this question was about approximations, not beta functions =)2012-01-15

3 Answers 3

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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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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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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.2012-01-15