How do you prove that ∫(e^(sec(x)))dx is impossible?

Question

Nice to see some helpful answers

mastap425: cos(1) is not 1, but this doesn't help, I know that the function e^(sec(x)) is not defined for all x, I need to know why the integral can't be done.

Sorry, maybe I can rephrase the question:

How do you prove that there is no function f(x) such that f'(x) = e^(sec(x))

Answer 1

That's not really a very precise statement of the problem.

Consider: on any closed interval [a, b] on which cos x is never 0, e^(sec x) is a continuous function and hence integrable. So ∫e^(sec x) dx is not impossible per se, over a defined interval on which cos x is never 0. Though calculating it might be a pain. ;-)

Now, to prove that we cannot calculate an improper integral for ∫e^(sec x) dx on any interval containing a point where cos x = 0. Obviously e^(sec x) is periodic, so we only need to consider one cycle, say [0, 2π], with critical points at π/2 and 3π/2. By symmetry of the cos function, we can get away with only looking at one of these, say π/2.

Consider the interval [π/2 - Δx, π/2). On this interval cos(x) is positive and decreasing (towards 0), so sec(x) is positive and increasing (towards ∞). So the minimum value for f(x) = e^(sec x) on this interval will occur at π/2 - Δx. We will use this to calculate a Riemann partial sum on this interval.

Now cos(π/2-Δx) = sin(Δx) ~ Δx - (Δx)^3/3! for Δx <<1.
So e^(sec (π/2-Δx)) ~ e^(1/Δx) to first order. The Riemann sum for this interval is therefore of the order Δx.e^(1/Δx). Let u = 1/Δx and we get e^u/u. As Δx -> 0+ we have u -> +∞ and the Riemann sum also goes to ∞. So the improper integral cannot be defined.

(Note that the other side of π/2 is fine, since the integrand goes to 0. A graph of e^sec x is quite an interesting one.)

Answer 2

What do you mean is impossible?
It's not impossible. It's true it jumps up and down, but that doesn't mean it doesn't work on a closed interval of -1 to 1 maybe?
It shoots up to infinity, so the area would be infinite. You didn't give us the max and mins for the integral.
By the way, cos(1) IS NOT 1 or 0 for those people above.

Answer 3

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Answer 4

if x is 1 then that would mean e^ (1/cos(1)) is equal to e^(1/0) therefore dividing by zero which you cannot do. Hope that helps!

Answer 5

It's impossible because you have to divide by zero, which is undefined.

Answer 6

i don do math unless its absolutely necessary but if it helps 9+9=18

Answer 7

sry my level of math is not that high ..but..everything is possible..so it is impossible that that is impossible ..dammit i just contradict myself mmmm ..