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In a proof to show that $\int_{0}^{1} f \left(\left\{1/x\right\}\right) \frac{ \mathrm{d}x}{1-x}=\int_{0}^{1} f(v) \frac{ \mathrm{d}v}{v}$, i found this line : $$\sum_{k=1}^{\infty}\int_{\frac{1}{k+1}}^{\frac{1}{k}} f \left(\left\{1/x\right\}\right) \frac{ \mathrm{d}x}{1-x} = \sum_{k=1}^{\infty} \int_{k}^{k+1} f \left(\left\{ u \right\}\right) \: \frac{\mathrm{d} u}{u(u-1)}$$

I'm having some trouble understanding the process $\int_{\frac{1}{k+1}}^{\frac{1}{k}} f \left(1/x\right) \frac{ \mathrm{d}x}{1-x}=\int_{k}^{k+1} f \left(u \right) \: \frac{\mathrm{d} u}{u(u-1)}$.

Using the formula $$\int_{\phi(a)}^{\phi(b)}f(x)\,\mathrm dx=\int_a^bf\big(\phi(t)\big)\phi^\prime(t)\,\mathrm dt.$$ with $\phi(x)=1/x$, i lack the $\phi'(x)$. How does it work ?

Hakim
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Hippalectryon
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  • $x=\frac{1}{t}, dx=-\frac{1}{t^2}dt$ – Shine Jul 31 '14 at 15:28
  • what do you mean by $\int_{0}^{1} f \left(\left{1/x\right}\right) \frac{ \mathrm{d}x}{1-x}=\int_{0}^{1} f(v) \frac{ \mathrm{d}v}{v}$, if it is correct, $f$ must be a special function. – Shine Jul 31 '14 at 15:31
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    @Hippalectryon (and Shine) Just to put the related hyperlink: http://math.stackexchange.com/questions/870680/another-integral-for-pi Thanks! – Olivier Oloa Jul 31 '14 at 16:46
  • @Shine it works for any integrable function if my proof is correct : http://mathb.in/18845 – Hippalectryon Jul 31 '14 at 17:05

2 Answers2

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Making the substitution $\frac{1}{x}=u$ you get $-\frac{1}{x^2}dx=du$; using the substitution it becomes $dx=-\frac{1}{u^2}du$. So you can rewrite the integral as $$-\int_{k+1}^k f(u)\frac{1}{1-\frac{1}{u}}\frac{du}{u^2}=\int_{k}^{k+1} f(u)\frac{du}{u(u-1)}$$

max
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Letting $$u=\frac{1}{x}$$

we have $$\frac{du}{dx}=\frac{-1}{x^2}$$

so that $$\frac{dx}{du}=-x^2 => dx = \frac{-1}{u^2}du$$

then just substitute $$x=\frac{1}{u}$$

to obtain the integral over u. The integration limits are then 1 to infinity over u, of course using the above subsitution, not 0 to 1.

Cheers,

aethelred_the_unwashed
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