dr. undefined Posted March 12, 2012 Share Posted March 12, 2012 Integration of e^sinx with respect to x. Do you think it ever terminates? Link to comment Share on other sites More sharing options...

DrRocket Posted March 12, 2012 Share Posted March 12, 2012 Integration of e^sinx with respect to x. Do you think it ever terminates? What in the world is that supposed to mean ? Link to comment Share on other sites More sharing options...

mississippichem Posted March 12, 2012 Share Posted March 12, 2012 (edited) 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 e^{1} 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, 2012 by mississippichem Link to comment Share on other sites More sharing options...

dr. undefined Posted March 15, 2012 Author Share Posted March 15, 2012 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 e^{1} 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? Link to comment Share on other sites More sharing options...

The Observer Posted March 17, 2012 Share Posted March 17, 2012 The integral from a to infinity does not exist if thats what you are asking. Link to comment Share on other sites More sharing options...

questionposter Posted March 22, 2012 Share Posted March 22, 2012 (edited) 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, 2012 by questionposter Link to comment Share on other sites More sharing options...

Shadow Posted March 24, 2012 Share Posted March 24, 2012 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. Link to comment Share on other sites More sharing options...

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