4
$\begingroup$

I proved that $$\int_{1}^\infty \left(\frac{1}{\lfloor x\rfloor}-\frac{1}{x}\right)\mathrm dx=\gamma.$$

I now have to deduce that $$1-\int_1^\infty \frac{x-\lfloor x\rfloor}{x^2}\mathrm dx=\gamma.$$

So what I want to prove is that $$\int_1^\infty \frac{x-\lfloor x\rfloor}{x^2}\mathrm dx=1-\gamma=1-\int_1^\infty \left(\frac{1}{\lfloor x\rfloor}-\frac{1}{x}\right)\mathrm dx=1-\int_1^\infty \frac{x-\lfloor x\rfloor}{x\lfloor x\rfloor}\mathrm dx.$$

To do this, I use the fact that $$1=\int_1^\infty \frac{1}{x^2}\mathrm dx$$ and thus, I get that $$1-\int_1^\infty \frac{x-\lfloor x\rfloor}{x\lfloor x\rfloor}\mathrm dx=\int_1^\infty \left(\frac{1}{x^2}+\frac{\lfloor x\rfloor-x}{x\lfloor x\rfloor}\right)\mathrm dx.$$

But I don't see how to deduce that $$\int_1^\infty \left(\frac{1}{x^2}+\frac{\lfloor x\rfloor-x}{x\lfloor x\rfloor}\right)\mathrm dx=\int_1^\infty \frac{x-\lfloor x\rfloor}{x^2}\mathrm dx.$$

I tried many manipulation, but it wasn't conclusif.

$\endgroup$

1 Answer 1

0
$\begingroup$

The $1/x$ terms cancel, so what you want to prove is

$$ \int_1^\infty\left(\frac1{\lfloor x\rfloor}-\frac{\lfloor x\rfloor}{x^2}\right)\mathrm dx=1\;. $$

Integrate by parts:

$$ \left[\frac{\lfloor x\rfloor}x\right]_{1\uparrow}^\infty+\int_{1\uparrow}^\infty\left(\frac1{\lfloor x\rfloor}-\frac{\lfloor x\rfloor'}x\right)\mathrm dx=1\;. $$

The first term is $1$, and in the integral both terms run through the positive integers, so they cancel.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.