For the following two problems, does the integral diverge or converge (if so, to what)
Integral of sinx/(cosx - 1) from -pi/2 to pi/2
Integral of e^x/x from 1 to infinity
Thanks!
Hint: this is an odd function integratd over a balanced interval
consider a slightly different problem: does the sum $\displaystyle \sum_{n = 1}^\infty \frac {e^n}n$ converge? what can you say about our integral if it does (or doesn't)Integral of e^x/x from 1 to infinity
So this integral diverges. So do you basically evaluate the given function taking into account the limits and then whatever happens to it happens to the area under the function?
Also, for the first problem, do you tell that it's odd simply by plugging in a test value? And then once you know it's odd, what does it mean that its evaluated over a symmetric interval (I have to be able to explain this tomorrow). Thanks so much for the help!
yes. at first i was trying to piggy back on the integral test, but i realize that is not applicable here, since we have an increasing function. thus, we can use an argument similar to what you just gave. the integral gives the net area under the curve, and we have a curve that is diverging to infinity, exponentially so; thus, the area under the curve (and hence the integral) will diverge as well.
a function $\displaystyle f(x)$ is odd, if $\displaystyle f(-x) = - f(x)$. when we integrate an odd function over a balanced interval, we get zero. you can look up why this is. the basic intuitive argument is that, again, the integral gives the net area under the curve. an odd function is symmetric about the origin, and so, our function is the same on both sides of the y-axis, except one side is flipped over and thus gives the "negative" area/integral of the other side. and they cancel outAlso, for the first problem, do you tell that it's odd simply by plugging in a test value? And then once you know it's odd, what does it mean that its evaluated over a symmetric interval (I have to be able to explain this tomorrow). Thanks so much for the help!