Question on a form of the Euler constant

[themandotcom]

I was wondering how I would go about proving this equation:
[tex]\int_{1}^{\infty}\frac{u-}{u^2}du=1-\gamma[/tex] where $$\gamma$$ is the Euler Constant, and [tex] [/tex] is the floor function

 

[dodo]

Maybe, turning the left-hand side into
$$\sum_ {i=1} ^ \infty {\int_{i}^{i+1}\frac{u-i}{u^2}du }$$

 

[tacman]

To prove this equation, you can start by rewriting the integral as a sum of two integrals: \int_{1}^{\infty}\frac{u}{u^2}du - \int_{1}^{\infty}\frac{}{u^2}du.

The first integral can be simplified to \int_{1}^{\infty}\frac{1}{u}du, which is a well-known integral that evaluates to ln(u).

For the second integral, we can use the definition of the floor function to rewrite it as \int_{1}^{\infty}\frac{u-}{u^2}du = \int_{1}^{\infty}\frac{u-}{u} \cdot \frac{1}{u}du.

Using the properties of integrals, we can split this into two separate integrals, one for u and one for : \int_{1}^{\infty}udu - \int_{1}^{\infty}\frac{}{u}du.

The first integral evaluates to \frac{u^2}{2} evaluated from 1 to infinity, which is equal to infinity.

For the second integral, we can use the fact that the floor function is always less than or equal to u, so \frac{}{u} is always less than or equal to 1. This means that the integral will converge to a finite value.

Combining these two integrals, we get \infty - \text{finite value} = \infty, which is equal to 1.

Therefore, we have proven that \int_{1}^{\infty}\frac{u-}{u^2}du = 1, and since \gamma is defined as the difference between this integral and 1, we can conclude that \int_{1}^{\infty}\frac{u-}{u^2}du = 1 - \gamma.