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.
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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 $2\log(x+1) - \log(x^2 - x + 1) \to 0$ .
Aryabhata
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See this answer: http://math.stackexchange.com/questions/34351/simpler-way-to-compute-a-definite-integral-without-resorting-to-partial-fraction/34362#34362 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 02 '12 at 22:27
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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 02 '12 at 22:45
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@Martin: You are welcome. I have added another method, which does use partial fractions. – Aryabhata Apr 02 '12 at 22:47
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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. $$\int_0^\infty \frac{1}{1+x^3} \ dx = \frac{\pi}{\sqrt{3}} + \frac{\arctan(1/\sqrt{3})}{\sqrt{3}} = \frac{2 \pi}{3 \sqrt{3}}$$
Robert Israel
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Thank you Robert. Nice solution by taking asymptotics. Little typo: in the last integral the upper bound is $\infty$ – Martin Apr 03 '12 at 18:39
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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.
