Integration of e^sinx with respect to x. Do you think it ever terminates?

Integration of e^sinx with respect to x. Do you think it ever terminates?

 

What in the world is that supposed to mean ?

Integration of e^sinx with respect to x. Do you think it ever terminates?

 

Expand [math] e^{\sin x} [/math] as a taylor series at [math] x_{0}=0[/math]. Integrate it between 0 and infinity. When you try to take the limit you'll see.

 

Also. You could just plot the function or think about the fact that the function oscillates sinusoidally between e^-1 and e¹ as 1 and -1 are both the extreme values of sin x.

 

This is all assuming that by "does it terminate?" you mean does the integral from 0 to infinity have a finite value?

Edited March 12, 201213 yr by mississippichem

Expand [math] e^{\sin x} [/math] as a taylor series at [math] x_{0}=0[/math]. Integrate it between 0 and infinity. When you try to take the limit you'll see.

 

Also. You could just plot the function or think about the fact that the function oscillates sinusoidally between e^-1 and e¹ as 1 and -1 are both the extreme values of sin x.

 

This is all assuming that by "does it terminate?" you mean does the integral from 0 to infinity have a finite value?

 

So it doesn't have a generalized answer for indefinite integrals?

The integral from a to infinity does not exist if thats what you are asking.

Doesn't that kind of make sense though? If you have an infinitely long function, then there's infinite area under it. I suppose there should be a nodal at infinity itself though, but that doesn't really mean anything in reality.

Edited March 22, 201213 yr by questionposter

That's not true. The function [math]\frac{1}{(n+1)^2}[/math] is "infinitely long" and the area under it (from 0 to infinity) sums up to one. For an example of a function whose integral from negative to positive infinity (can one say "over the real numbers"?) is finite, have a look at the Gaussian integral.