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

Question

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.

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

Answer 1

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 $\frac{\pi}{2\sqrt{2}} \approx 1.11072$.

Answer 2

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.

Answer 3

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.