Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

4 Answers 4

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 $2\log(x+1) - \log(x^2 - x + 1) \to 0$ .

share|improve this answer
    
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)). –  Aryabhata Apr 2 '12 at 22:27
    
Thank you Aryabhata. This is a very elegant way of evaluation by reducing it to the $\int_0^\infty \frac{dx}{x^2 - x + 1}$. –  Martin Apr 2 '12 at 22:45
    
@Martin: You are welcome. I have added another method, which does use partial fractions. –  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. $$\int_0^\infty \frac{1}{1+x^3} \ dx = \frac{\pi}{\sqrt{3}} + \frac{\arctan(1/\sqrt{3})}{\sqrt{3}} = \frac{2 \pi}{3 \sqrt{3}}$$

share|improve this answer
    
Thank you Robert. Nice solution by taking asymptotics. Little typo: in the last integral the upper bound is $\infty$ –  Martin Apr 3 '12 at 18:39
    
Thanks for spotting that, fixed it. –  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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.