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Compute integral $$\int_0^{\infty} \frac{dx}{1+x^3}$$.

I used the formula from complex variables by fisher (2.6 formula (9)) that $$\int_0^{\infty} \frac{dx}{1+x^a}=\frac {\pi} {a\sin(\pi/a)}$$ This means $$\int_0^{\infty} \frac{dx}{1+x^3}=\frac {\pi} {3\sin(\pi/3)}$$ Is that simply the answer or do I need to expand it to find all concrete values?

Additionally, from this answer here, the improper integral gives a different result:

$$-\frac{1}{6} \ln |x^2-x+1|+\frac{1}{\sqrt{3}} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|$$

is this result equal to the one I got above?

metamorphy
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james black
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    It's not that it gave a different result. You simply calculated the indefinite integral $F(x) = \int 1/1+x^3,dx$. What you need to do now is consider $\lim_{b\to\infty} F(b) - F(0)$ to match the two different answers. – Hyperion Jan 23 '21 at 21:39
  • oh so the other answer is an indefinite integral--what i got is a simple definite integral? but if i got the definite integral, why do i still need to consider the limit? please forgive me but I am new to complex integral so please guide me a bit on calculating the limit – james black Jan 23 '21 at 21:42
  • Several answers in the link you gave give a value for the integral of $\frac{2\pi}{3\sqrt{3}}.\ $ Also, $\ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}.\ $ Therefore the answer your formula gives is $\large{\frac{\pi}{3\sin\left(\frac{\pi}{3}\right) } = \frac{\pi}{3\frac{\sqrt{3}}{2} } = \frac{2\pi}{3\sqrt{3}}}.\ $ This matches the other answers, so yes, you can be confident that that is the correct answer. I'm not sure what you mean by, " or do I need to expand it to find all concrete values?" – Adam Rubinson Jan 23 '21 at 21:43
  • understood thank you my confusion is basically "why is $\frac {\pi} {3sin(\pi/3)}$ not the final answer? and why do we need to proceed to calculate the lim of F(b)-f(0)"? or i am confused and $\frac {\pi} {3sin(\pi/3)}$ is the final answer--it is just the quesiton is literally one step and seemed too easy... – james black Jan 23 '21 at 21:49
  • @jamesblack perhaps the question wanted you to find this answer without consulting any tables or integral solvers. Would you like me to show how to find the answer for yourself? – A-Level Student Jan 23 '21 at 22:04

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Next, simplify $$ F(x)=-\frac{1}{6}\ln|x^2-x+1|+\frac{1}{\sqrt{3}}\arctan{\frac{2x-1}{\sqrt{3}}}+\frac{1}{3}\ln|x+1| $$ $$ =\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|-\frac{1}{3}\ln\sqrt{|x^2-x+1|} $$ $$ =\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln\left(\frac{|x+1|}{\sqrt{|x^2-x+1|}}\right). $$ Then $$\int_0^\infty \frac{dx}{1+x^3}=\lim_{X\rightarrow\infty}F(X)-F(0).$$ Compute the limit, and you are done.

John Peter
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