Integrating the fractional part of a logarithm function

My first fractional part integral and the solution development is pretty elegant indeed.

[manstuckinabox]

I like how the challanging part of these integrals isn't solving the integral, it's to deal with the sums afterwards, like bruh, I came here to integrate not get disintegrated by freaking sums. Great videa as always meine freunde.

[BadlyOrganisedGenius]

Another way to do this would be the substitution u = e*x
Because then x = u/e ==> {-ln(x)} = {-ln(u/e)} = {1-ln(u)} = {-ln(u)}
So I = (1/e)*(Integral from 0 to e of {-ln(u)} ) = (1/e)*( I + (Integral from 1 to e of 1-ln(u) ) which is easy to calculate, leaving us with an equation for I.

Great video!

[angus]

using basically the exact same method as in this video, you can do the integral from 0 to 1 of {1/x}, and part of it reduces to the definition of the euler mascheroni constant, overall giving an answer of 1 - gamma

[dhyanlaad]

I was able to solve this one before looking at your solution, outstanding video as usual

[MrWael]

Very good solution. Really impressive trics.

[driesvanheeswijk]

Man I sure do love my daily dose of Maths 505.

I have never had to do weird integrals like these but they look really fun! Do they ever come up in a standard Mathematics bachelor?

[krisbrandenberger]

In the telescoping sum, the n+1 in the exponent should be in parentheses after the telescoping action.

[kewalmer]

nice question, I did it by just keeping it as the logarithmic function and writing {x}=x-[x]

[FRANKONATOR]

Anything with fractional part, floor, ceiling, etc., is going to be interesting

[ahmedhamdi]

Try lnx instead of -lnx in the integrand