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I'd like your help with see why does $$\int^1_0 \frac{1}{e^t} \; dt $$ converge?

As I can see this it is suppose to be: $$\int^1_0 \frac{1}{e^t}\;dt=|_{0}^{1}\frac{e^{-t+1}}{-t+1}=-\frac{e^0}{0}+\frac{e}{1}=-\infty+e=-\infty$$

Thanks a lot?

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What? No! $\int e^{-t} \mathrm dt$ shouldn't give that result... –  J. M. Dec 6 '11 at 11:26
    
$\dfrac{e^{-t+1}}{-t+1}$ doesn't differentiate to $\dfrac1{e^t}$, for one. –  J. M. Dec 6 '11 at 11:28
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$\int e^{kx}\,dx={e^{kx}\over k}+C$... –  David Mitra Dec 6 '11 at 11:32
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@Jozef: Making a mistake is not a reason for deleting a question. What if someone else makes the same mistake and looks for the answer? We want this site to be able to help everyone. –  Zev Chonoles Dec 6 '11 at 11:38
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@JackManey: Yeah, I got it.. –  Jozef Dec 7 '11 at 8:30
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6 Answers 6

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The integration you have shown is not correct. Exponential functions are quite different to power functions and have very different antiderivatives. The correct integration is as follows:

$$\int_0^1 e^{-t} \, dt = -e^{-t}|_0^1 = 1-\frac{1}{e}.$$

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$$ \int_0^1 e^{-t} = [-e^{-t}]_0^1 = -\frac{1}{e} + 1$$

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I assume you are integrating over the $t$ variable. $1/e^t$ is a continuous function, and you are integrating over a bounded interval, so the integral is well defined. An antiderivative of $1/e^t=e^{-t}$ is equal to $-e^{-t}$. So the integral equals $1-1/e$

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Your calculation is incorrect. Instead, you should have $$\int_0^1\frac{1}{e^t}dt=\int_0^1e^{-t}dt=-e^{-t}\big|_0^1=(-e^{-1}-(-e^{0}))=1-\frac{1}{e}$$ You got mixed up with the rule for powers, $$\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$$ but for exponentials we have $$\int e^x\,dx=e^x+C$$

You can also see that $e^{-t}$ is bounded above by the constant function $1$ on the interval $[0,1]$:

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ enter image description here

so that the area underneath the curve $e^{-t}$ from $0$ to $1$ has to be less than the area underneath the curve $1$ from $0$ to $1$ (which is $1$). So this picture tells you that the value of $\int_0^1\frac{1}{e^t}dt$ is less than $1$, which we've confirmed by showing that is in fact equal to $1-\frac{1}{e}$.

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Others have given a correct way to evaluate the integral. But even without evaluating the integral, you can see that it converges. After all, how can an integral fail to converge? It can fail to converge if one or both of the limits of integration is infinite, e.g., $\int_1^{\infty}x^{-1}\,dx$ fails to converge. It can fail to converge if the integrand is unbounded, e.g., $\int_0^1x^{-1}\,dx$ fails to converge. But if the limits of integration are bounded, and the integrand is bounded, the integral has no way to diverge - it must converge.

In your example, the limits of integration are bounded, and the integrand satisfies $1/e\le e^{-t}\le1$ for $t$ in the interval of integration ($0\le t\le1$), so convergence is immediate.

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$@$Gerry: some kind of continuity hypothesis in your first paragraph would be nice. –  Pete L. Clark Dec 6 '11 at 11:56
    
@Pete, shush! At this level of mathematics, all functions are continuous, or near enough as to make no difference. Anyway, the kind of function you're thinking of, I would say the integral doesn't exist, but I'm not sure I would say it doesn't converge; its problems are deeper than those arising from improper integrals. –  Gerry Myerson Dec 6 '11 at 12:20
    
I have taught calculus at many levels, including to freshman who will absolutely not be taking any further math courses, but at none of these levels is "all functions are continuous, or near enough as to make no difference" a true statement. I agree that underscoring the need for continuity is probably a poor pedagogical choice in this particular instance...but you shouldn't say things that are false, especially in an environment where lots of other people will be hearing or seeing them.... –  Pete L. Clark Dec 6 '11 at 12:45
    
..."But if the limits of integration are bounded, and the integrand is bounded, the integral has no way to diverge - it must converge." As you well know, this is simply not true (or perhaps not meaningful if we are not sure about the distinction between convergence and existence, but in every instance I can think of "convergence" means "some limit exists", and here we have a kind of limit that need not exist). I speak from experience when I say that using the word "continuous" will not rattle calculus students...as long as you don't ask them to define it or do anything with it. –  Pete L. Clark Dec 6 '11 at 12:50
    
@Pete, I just finished teaching a semester of Calculus, and I did it without using the word "limit", without using the word "continuous", without making use of any function that wasn't continuous wherever it was defined. Had I decided to mention a discontinuous function, it would have been something like $-1$ for $x\le0$, $1$ for $x\gt0$, which is (Riemann-)integrable on bounded intervals. I'll stand by my statement - you just haven't taught calculus at the level of Math 130 at Macquarie. –  Gerry Myerson Dec 7 '11 at 3:35
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If you mean: How do you know the integral exists and has a finite value, the answer is simply that it's the integral of a continuous function over a bounded interval.

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