Integration with upper limit infinity

[muso07]

Homework Statement

$$\stackrel{\infty}{0}\int2e^{ky}dy=3/2$$

Homework Equations

The Attempt at a Solution

I got up to:
$$\stackrel{lim}{x\rightarrow\infty}\[[e^{ky}]^{x}_{0}=3k/4$$
$$\stackrel{lim}{x\rightarrow\infty}\[[e^{kx}]=\frac{3k+4}{4}$$

I have no idea how to work that out. Any help will be much appreciated.

 

[Dick]

Can you show us how you got those crazy formulas?

 

[muso07]

I went

$$\frac{2}{k}\stackrel{\infty}{0}\int(k)e^{ky}dy=3/2$$
$$\stackrel{\infty}{0}\int(k)e^{ky}dy=3k/4$$

^ That's the integral from 0 to infinity, I couldn't figure out how to make it look nice.

 

[Dick]

Ok, the integral from 0 to x of 2*e^(ky)dy is (2/k)(exp(ky)) from y=0 to y=x. Which is limit (2/k)*(exp(k*x)-1) as x->infinity. Better assume k<0 or it won't converge. What's that limit? Equate that to 3/2. What's k?

 

[Mark44]

I went

$$\frac{2}{k}\stackrel{\infty}{0}\int(k)e^{ky}dy=3/2$$
$$\stackrel{\infty}{0}\int(k)e^{ky}dy=3k/4$$

^ That's the integral from 0 to infinity, I couldn't figure out how to make it look nice.

$$\frac{2}{k}\int_0^{\infty}(k)e^{ky}dy$$
$$=\frac{2}{k}\lim_{b \rightarrow \infty}\int_0^{b}e^{ky}kdy$$
$$=\frac{2}{k}\left[\lim_{b \rightarrow \infty}e^{ky} \right]_{y = 0}^{b}$$
In the 2nd line above, you used the substitution u = ky, du = kdy. Can you take it from here?

Note: I made it "look nice." You can see my LaTeX script by clicking on any of the lines.

 

[muso07]

$$\left[\lim_{x \rightarrow \infty}e^{kx} \right]=\frac{3k}{4}+1$$

This is what I'm up to (as in the OP), and I get everything up to here, but it's the limit I'm having trouble with.

 

[Mark44]

How on Earth did you get that? I'm especially curious to learn where the 3 came from.

Do you know what the graph of y = 2e^kx looks like. Your original integral can be interpreted as the area between the graph of this function and the x-axis, for x >= 0. The value you get depends on k, which you didn't specify in the OP.

 

[muso07]

The 3 comes from the original problem, where the RHS was 3/2.

But I've now realized I wrote the problem down wrong, the RHS is 1/2 not 3/2.

I'm meant to be solving for k, which is why I didn't specify the value for k.

So it's really meant to be
$$<br /> \left[\lim_{x \rightarrow \infty}e^{kx} \right]=\frac{k}{4}+1<br />$$

I'm not trying to find what the integral equals, I'm solving for k but I'm still confused about how to get that limit out.

 

[Mark44]

So is this the problem?
Find k so that
$$\int_0^{\infty} 2 e^{ky} dy~=~ \frac{1}{2}$$
This is equivalent to
$$\int_0^{\infty} e^{ky} dy~=~ \frac{1}{4}$$

To solve for k, you need to carry out the integration, and since the integral is improper (one limit is infinity), you need to evaluate this integral.
$$\lim_{b \rightarrow \infty} \int_0^{b} e^{ky} dy~=~ \frac{1}{4}$$

For some values of k, the limit will not exist, because e^(ky) grows too large. You need to determine conditions on k so that the limit exists, and therefore that the integral is a finite number. See my work in post 5 of this thread.

 

[muso07]

Ok, I think I got it.

If k<0, then that limit is zero, which means k=-4?

 

[Mark44]

Yes.

 

[muso07]

Sweet, thanks. :)