# Proving convergence for integral

1. Nov 16, 2015

### zebo

1. The problem statement, all variables and given/known data
Prove that for every a ∈ ℝ+ the following improper integrals are convergent and measure its value.

∫a∞exp(-at)dt
Edited by mentor: $\int_a^{\infty} e^{-at} dt$

∫1∞exp(-2at)dt
Edited by mentor: $\int_1^{\infty} e^{-2at} dt$
3. The attempt at a solution

For the first integral i get -1/t+exp(t^2)+1/aexp(a^2) which for t going to infinity is convergent with the value 1/aexp(a^2)

For the second integral i get that it converges towards 1/2aexp(2a) for t going to infinity.

My issue is, that i am not quite sure how to explain, that this shows that for every a > 0 the integrals converges?

Last edited by a moderator: Nov 16, 2015
2. Nov 16, 2015

### zebo

Dont know what went wrong when i posted, but the integrals are meant to be read as from a to ∞ and 1 to ∞ for the functions exp(-at) and exp(-2at)

3. Nov 16, 2015

### LCKurtz

You need to show your work. For example, for the first one you need to show
$$\lim_{b\to \infty}\int_a^b e^{-at}~dt$$ is finite and give its value. And check your antiderivative. And did you mean to have $a$ both in the lower limit and the integrand?

4. Nov 16, 2015

### zebo

Yes a is both in the lower limit and in the integrand in the assignment, was confused by this at first as well. I assign t > a in the first one, and t > 1 in the second one, and then i let t go towards infinity to show that they are convergent with the values i found when t goes towards infinity. Is this enough though, to conclude that for every a > 0 they are convergent?

Also, the last part of the assignment is to solve the equation: I1 = I2 (I1 is the first integral and I2 is the second integral)

5. Nov 16, 2015

### LCKurtz

Your first one is wrong and we have nothing to discuss until you show your work. It doesn't help us for you to just tell us "what you got". And even then, you need to put parentheses in the correct places for us to even know what you did get.

Last edited: Nov 16, 2015
6. Nov 16, 2015

### MidgetDwarf

In your calculus book, this topic is called improper integrals. There are generally 2 types of improper integrals for your case. I can help you, but like others have said, you have to show some work.

7. Nov 16, 2015

### zebo

The first one, i assign t > a as the upper limit. When i solve it i get (-1/t)*(1/e^(t^2)) - (-1/a)*(1/e^(a^2)) = 1/(a*e^(a^2)) - 1/(t*e^(t^2)) . For t → ∞, the integral → 1/(a*e^(a^2))

8. Nov 16, 2015

### zebo

I dont know what you mean? I integrate it and get -1/a * exp(-at) = - 1/(a*exp(at)) which gives me the above answer when i use the upper and lower limit.

9. Nov 16, 2015

### Ray Vickson

But in post #1 you wrote "For the first integral i get -1/t+exp(t^2)+1/aexp(a^2)". I copied/pasted this directly from your message!

Last edited: Nov 16, 2015
10. Nov 16, 2015

### zebo

Oh yeah, went alittle fast from my paper to here, sorry about that i ofcourse meant -1/(t*exp(t^2)) not 1/t+exp(t^2).

I am heading to bed, but i will work on it tomorrow. It is not so much the solving of the integrals, but more the understanding behind why this shows that for all positive values of a this shows that the integrals are convergent with above mentioned values. If the above shows the solution in which a takes part, then how do i explain / understand that this shows that for all positive values of a it is convergent?

On another note, i am asked to solve the equation 1/(a*exp(a^2)) = 1/(2a*exp(2a)) and i cant quite wrap my head around how to begin.

Thx in advance for any / all help and sorry if i am coming off as a guy who just wants you to solve my stuff, since this is not the case. If you have any link to someplace where i can understand this better it would be greatly appreciated.