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Where can I find the integrals (the convergence or divergence of which is well known) to use in the comparison test and limit comparison test for convergence of integrals?

I know that

$$ \int_0^1 x^\alpha\mathrm{d}x = \begin{cases} \frac1{\alpha + 1} &\alpha\in(-1, \infty), \\[.6em] \infty &\alpha\in(-\infty, -1), \\[.9em] \infty &\alpha=-1 \end{cases} $$

and

$$ \int_1^\infty x^\alpha\mathrm{d}x = \begin{cases} \infty &\alpha\in (-1, \infty), \\[.6em] -\frac1{\alpha+1} &\alpha\in (-\infty, -1), \\[.7em] \infty &\alpha=-1. \end{cases} $$

When proving the convergence of a given integral, I often split the interval on which I integrate, ie.

$$ \int_0^1 f(x)\mathrm{d}x = \int_0^\epsilon f(x)\mathrm{d}x + \int_\epsilon^1 f(x)\mathrm{d}x, \hspace{2em}\epsilon \in (0, 1). $$

Not only in this case would I like to compare against a much richer set of well known integrals. That is, I am looking for something similar to "basic series" used to conclude on convergence and divergence of a series.

David
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    Not sure this is what you're looking for, but you can use the integrals in the second part of this answer has reference integrals for comparison. – Git Gud Sep 08 '13 at 14:46

1 Answers1

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Perhaps you'd like to add $$\int_2^\infty \frac{dx}{x(\ln x)^\alpha}$$ to your list. Of course, with a change of variables it reverts to your list. In addition, $$\int_0^\infty e^{-x}\,dx$$ is always useful, for example, to consider $\int_0^\infty e^{-x^2}\,dx$, but it too can be deduced by direct comparison, for example, with $\int_1^\infty x^{-2}\,dx$.

Ted Shifrin
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  • Nice tip, thanks. I just realized this - how do I ever use a comparison test if I split the integral the way I have shown in my question (for some arbitrary $\epsilon$)? – David Sep 08 '13 at 15:01
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    In my experience, the splitting is usually more needed when it's an integral from $0$ to $\infty$, perhaps improper on both ends, perhaps not. For example, to do $\int_0^\infty e^{-x^2},dx$, I'd observe that there's no problem on $[0,1]$ and then compare with $\int_1^\infty e^{-x},dx$ on $[1,\infty)$. – Ted Shifrin Sep 08 '13 at 15:46
  • I have something like $\int_0^1 \frac{\ln(1 - x)\sqrt{x - x^2}}{\sin(\pi x^2)}\mathrm{d}x$ in my mind. I found it useful to split this into two integrals with one "proper end" each. I have found nice integral to compare with around zero (in the form $\int x^\alpha\mathrm{d}x$) - but I do not know whether it converges on $(0, \epsilon]$. Hence the question. – David Sep 08 '13 at 16:15
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    Perhaps this should have been your original question :) I would recommend working with the Taylor expansion—where meaningful—near each of $0$ and $1$. Since you raise the question about $0$, let's see: $\ln(1-x)\approx -x+o(x)$, $\sin(\pi x^2)\approx \pi x^2$, and $\sqrt{x-x^2}\approx\sqrt x$, so the integrand looks like $\dfrac{-x\sqrt x}{\pi x^2}\approx \dfrac1{\pi}\dfrac1{\sqrt x}$ and hence the integral converges. (You can put in the error terms and use the standard trick of writing $\dfrac1{\sin u} = \dfrac1{u(1-u^2/6+\dots)} = \dfrac1u(1+u^2/6+\dots)$ to be more rigorous.) – Ted Shifrin Sep 08 '13 at 16:22
  • Thank you, Ted. I have got the same integral to compare to (I worked it out looking at how those functions behave, not strictly from Taylor's expansion). You touch the very problem I have. How do you conclude that the integral converges? You cannot (at least I think so) use the limit comparison test. From what I know, you need the function to be either continuous at $(0, 1]$ or $[0, 1)$, but it is not defined for neither $x = 0$ nor $x = 1$. You are right, I should have asked directly. – David Sep 08 '13 at 16:57
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    Just do limit comparison on $(0,\epsilon)$ and $[1-\delta,1)$ separately. The error term's being $o(x)$ tells us exactly that limit comparison succeeds. – Ted Shifrin Sep 08 '13 at 17:21
  • I hope you don't mind me asking once more. I will try asking from the other side. We agreed that the best we can do is to compare with $\frac1{\pi\sqrt{x}}$. How do we know this is a convergent integral on the interval $(0, \epsilon]$? It is convergent on $(0, 1)$ for sure, as I wrote in the question. But why on $(0, \epsilon]$? I would really like to get this. – David Sep 08 '13 at 17:48
  • I think I found my error. I got mixed between Riemann and Newton integral. I got it! Thank you for help. – David Sep 08 '13 at 18:04