The integral $$\int_0^\infty \frac{dx}{1 + x^4} = \frac{\pi}{2\sqrt2}$$ can be evaluated both by a complex method (residues) and by a real method (partial fraction decomposition). The complex method works also for the integral $$\int_0^\infty \frac{dx}{1 + x^3} = \frac{2\pi}{3\sqrt3}$$ but partial fraction decomposition does not give convergent integrals. I would like to know if there is some real method for evaluating this last integral.


Make the substitution $x = \frac{1}{t}$ and you get

$$ \int_{0}^{\infty} \frac{t}{1+t^3} \text{d}t$$

Write the one you want as

$$ \int_{0}^{\infty} \frac{1}{1+t^3} \text{d}t$$

Now you can add both and cancel that pesky $1+t$ factor.

btw, a straightforward approach using partial fractions also works.

You consider

$$F(x) = \int_{0}^{x} \frac{1}{1+t^3} \text{d}t$$

Using partial fractions you can find that (I used Wolfram Alpha, I admit)

$$F(x) = \frac{1}{6}\left(2\log(x+1) - \log(x^2 - x -1) + 2\sqrt{3} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)\right) + \frac{\pi}{6\sqrt{3}}$$

Now as $x \to \infty$, we have that $2\log(x+1) - \log(x^2 - x + 1) \to 0$ .

  • $\begingroup$ See this answer: math.stackexchange.com/questions/34351/… to see that you can also evaluate the integral in that question to find your answer here, and then apply the other answers to that question (in particular, Eric's answer(s)). $\endgroup$ – Aryabhata Apr 2 '12 at 22:27
  • $\begingroup$ Thank you Aryabhata. This is a very elegant way of evaluation by reducing it to the $\int_0^\infty \frac{dx}{x^2 - x + 1}$. $\endgroup$ – Martin Apr 2 '12 at 22:45
  • $\begingroup$ @Martin: You are welcome. I have added another method, which does use partial fractions. $\endgroup$ – Aryabhata Apr 2 '12 at 22:47

Note that for $a > 0$, $$\int_0^N \frac{1}{x+a}\ dx = \ln(N+a) - \ln(a) = \ln(N) - \ln(a) + o(1)\ \text{as} \ N \to \infty$$ while $$\eqalign{\int_0^N \frac{x+a}{(x+a)^2 + b^2}\ dx &= \frac{1}{2} \left(\ln((N+a)^2+b^2) - \ln(a^2+b^2)\right)\cr &= \ln(N) - \ln(a^2+b^2) + o(1) \ \text{as} \ N \to \infty\cr}$$ and (if $b > 0$) $$ \eqalign{\int_0^N \frac{1}{(x+a)^2+b^2}\ dx = \frac{\arctan\left(\frac{N+a}{b}\right) - \arctan\left(\frac{a}{b}\right)}{b} = \frac{\pi}{2b} - \frac{\arctan\left(\frac{a}{b}\right)}{b} + o(1) \ \text{as} \ N \to \infty\cr}$$ In particular, from the partial fraction decomposition $$ \frac{1}{1+x^3} = \frac{1/3}{x+1} + \frac{(2-x)/3}{x^2 - x + 1} = \frac{1/3}{x+1} + \frac{1/2}{(x-1/2)^2+3/4} - \frac{(x-1/2)/3}{(x-1/2)^2 + 3/4}$$ you get $$ \int_0^N \frac{1}{1+x^3} \ dx = \frac{\ln(N) - \ln(1))}{3} + \frac{\pi/2 + \arctan(1/\sqrt{3})}{\sqrt{3}} - \frac{\ln(N) - \ln((1/2)^2 + 3/4)}{3} + o(1)$$ i.e. $$\int_0^\infty \frac{1}{1+x^3} \ dx = \frac{\pi}{\sqrt{3}} + \frac{\arctan(1/\sqrt{3})}{\sqrt{3}} = \frac{2 \pi}{3 \sqrt{3}}$$

  • $\begingroup$ Thank you Robert. Nice solution by taking asymptotics. Little typo: in the last integral the upper bound is $\infty$ $\endgroup$ – Martin Apr 3 '12 at 18:39
  • $\begingroup$ Thanks for spotting that, fixed it. $\endgroup$ – Robert Israel Apr 3 '12 at 21:01

For what is worth:

Your integral evaluates in terms of the sine function:

$$\int\limits_0^\infty \frac{1}{1+x^a}=\frac{\pi}{a}\sec\frac{\pi}{a}$$

refer to this question and the link in it.


I would like to know if there is some real method for evaluating this last integral.

Actually, all integrals of the form $\displaystyle\int_0^\infty\frac{x^n}{1+x^m}dx$ can be solved by substituting $t=\dfrac1{1+x^m}$ , and then recognizing the expression of the beta function in the new integral, which can be written as a product of gamma functions. Then we use the reflection formula in order to finally arrive at the desired result, $I=\dfrac\pi m\cdot\csc\left[(n+1)\dfrac\pi m\right]$ — See my answer here for more information.


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