How Many Abelian Groups of Given Order

[darkchild]

Integrating Floor Function

Homework Statement

If $$\lfloor{x}\rfloor$$ denotes the greatest integer not exceeding x, then $$\int_{0}^{\infty}\lfloor{x}\rfloor e^{-x}dx=$$

Homework Equations

none

The Attempt at a Solution

I don't know how to start this problem. At first, I tried bringing the floor function outside of the integral and using a summation because it only takes on discrete values, then I realized that those discrete values get multiplied by the exponential infinitely many times as x varies from any integer to the integer that is one greater.

I've tried the product rule. I looked up both the derivative and the indefinite integral of the floor function. If I use $$u=\lfloor{x}\rfloor$$, du=0 and I get back the original integral. If I use $$dv=\lfloor{x}\rfloor$$, the integral just gets more complicated.

 

[Dick]

You aren't going to have much luck with the usual analytic techniques. Integrate it over the interval [n,n+1] for n an integer. So the floor function has a definite value. Then sum over n.

 

[darkchild]

You aren't going to have much luck with the usual analytic techniques. Integrate it over the interval [n,n+1] for n an integer. So the floor function has a definite value. Then sum over n.

The problem I'm having is that I can't integrate this at all. I'm not even at the step where I fiddle with the limits of integration/take sums yet.

 

[Dick]

The problem I'm having is that I can't integrate this at all. I'm not even at the step where I fiddle with the limits of integration/take sums yet.

You can't integrate a function like f(x)=n*e^(-x) where n is a constant? What's the integral for x=2 to x=3 of floor(x)*e^(-x)? Where I'm writing 'floor' for the function instead of the funny brackets.