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I would like to prove the convergence of the Newton integral

$$\int_0^1 f(x)\mathrm{d}x =\int_0^1 \frac{\sqrt{x-x^2}\ln(1-x)}{\sin{\pi x^2}} \mathrm{d}x.$$

I split this into two integrals $\displaystyle\int_0^\epsilon f(x)\mathrm{d}x$ and $\displaystyle\int_\epsilon^1 f(x)\mathrm{d}x$. It is easy to show that the integral is convergent on $(0, \epsilon]$ by limit comparison with $\displaystyle\int_0^\epsilon \frac{\sqrt{x}x}{\pi x^2}\mathrm{d}x$.

But I cannot find anything to compare with around $x = 1$ on $[\epsilon, 1)$.

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2 Answers 2

up vote 3 down vote accepted

If you write $x = 1 - \delta$, you obtain

$$\int_0^{1-\varepsilon} \frac{\sqrt{(1-\delta)(1-(1-\delta))}\,\ln \delta}{\sin \bigl(\pi(1-\delta)^2\bigr)}\, d\delta = \int_0^{1-\varepsilon} \frac{\sqrt{\delta(1-\delta)}\,\ln \delta}{\sin \bigl(\pi(2\delta-\delta^2)\bigr)}\,d\delta,$$

and the integrand of that can be compared to the harmless

$$\frac{\sqrt{x}\,\ln x}{2\pi x}.$$

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Thank you, I will look into it. From now on, I will strictly refer to simple mathematical objects as to "harmless" ones. –  David Čepelík Sep 9 '13 at 13:00
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Thanks, @Did. Better that way round, less room for misunderstanding. –  Daniel Fischer Sep 9 '13 at 13:16
    
If I understand this correctly, what you propose is substituting the troublesome $1 - x = \delta$. The interval of integration changes accordingly. (I assume that $\mathrm{d}x = -\mathrm{d}\delta$, where did the $-1$ go?) Then you compare the obtained integral to an expression containing the original variable $x$ again; shall I substitute here, too? I assume this is an obvious step which I am not yet familiar with. –  David Čepelík Sep 9 '13 at 13:18
    
Exactly. And a nice solution, if you ask me. (I usually post a comment saying "if you do not like the modifications, just reverse them" but in this case, for some reason it seems I forgot.) –  Did Sep 9 '13 at 13:19
    
@DavidČepelík The $-1$ went into the change of order of integration boundaries, $\int_\varepsilon^1 dx\leadsto\int_{1-\varepsilon}^0d(1-\delta)\leadsto \int_0^{1-\varepsilon}d\delta$. The $x$ in the function to compare to is a generic variable, you can also call it $y$, $\text{Fred}$, or $\delta$. –  Daniel Fischer Sep 9 '13 at 13:22
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Regarding to Quotient test for integrals with non-negative integrands, we have $$\lim_{x\to 1^-}{(1-x)^{0.6}|f(x)|}=0<\infty$$ so the integral $$\int_{\epsilon>0}^1|f(x)|dx$$ converges

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Needs another TU! +1 –  amWhy Sep 10 '13 at 1:03
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