MHB Exam Prep Part 2: Resolving Complex Integrals

[bmanmcfly]

Now that I'm finally getting the hang of some of these integration techniques with complex functions, I've come across another hole. In my game here.

That comes with the concept of resolving the definite integral of these complex functions.
What happens is that they fluctuate and so can create issues where the y would cross the x-axis and so needs to be handled differently.

So, is there a way, short of "feeling" to know where the zeroes are going to be? Maybe some trick with Trig and log functions?

Next, on the same vein, how do you handle definitive integrals where on of the extremes lands on a limit of a portion of the function...
from practice exercises, I arrived at the solution:
[math]=4(\sin(\frac{\pi}{2})+\ln(\abs(\sin(\frac{\pi}{2})))-(\sin(0)+\ln(\abs(\sin(0)))))[/math]
(where abs is supposed to mean absolute value...

Thanks in advance for the help.

 

[MarkFL]

I take it you were asked to evaluate:

$$I=4\int_0^{\frac{\pi}{2}}\cos(x)+\cot(x)\,dx$$

This is an improper integral, and I would write:

$$I=4\lim_{t\to0}\left(\int_t^{\frac{\pi}{2}}\cos(x)+\cot(x)\,dx \right)$$

$$I=4\lim_{t\to0}\left(1+\ln\left|\csc(t) \right| \right)=\infty$$

We find the integral does not converge.

 

[bmanmcfly]

I take it you were asked to evaluate:

$$I=4\int_0^{\frac{\pi}{2}}\cos(x)+\cot(x)\,dx$$

This is an improper integral, and I would write:

$$I=4\lim_{t\to0}\left(\int_t^{\frac{\pi}{2}}\cos(x)+\cot(x)\,dx \right)$$

$$I=4\lim_{t\to0}\left(1+\ln\left|\csc(t) \right| \right)=\infty$$

We find the integral does not converge.

Yes, that was what the equation simplified as... But you lost me a bit here... I get the putting this in terms of the limit, but how does
$$I=4\lim_{t\to0}\left(\int_t^{\frac{\pi}{2}}\cos(x)+\cot(x)\,dx \right)$$
lead to
$$I=4\lim_{t\to0}\left(1+\ln\left|\csc(t) \right| \right)=\infty$$
For some reason, I'm just not seeing it at the moment.
I get the finding the limit, was that because you used the definition of the limits ?
I'm just confused since this is the first time I've seen this method used. Did you just divid by cos?

But in a more general sense, is there a system to determine zeros in a complex equation (trig and log, whereas if you have a sin*cos situation or whatever?

Ik ow each cycle is 2pi, but that doesn't necessarily mean that the function will cross to zero. The closest I've been doing is just by charting a few key points and guessing the zeros.

 

[MarkFL]

I was a bit sloppy with my notation. Let's go back to:

$$I=4\lim_{t\to0}\left(\int_t^{\frac{\pi}{2}}\cos(x)+\cot(x)\,dx \right)$$

Now we know:

$$\int_a^b \cos(x)\,dx=\sin(b)-\sin(a)$$.

We also know that:

$$\int_a^b\cot(x)\,dx=\ln|sin(b)|-\ln|\sin(a)|=\ln\left|\frac{\sin(b)}{\sin(a)} \right|$$

So integrating term by term, we find:

$$I=\int_0^{\frac{\pi}{2}}\cos(x)+\cot(x)\,dx=\left(\sin\left(\frac{\pi}{2} \right)-\sin(t) \right)+\left(\ln\left|\frac{\sin\left(\frac{\pi}{2} \right)}{\sin(t)} \right| \right)=$$

$$1-\sin(t)+\ln\left|\frac{1}{\sin(t)} \right|=1-\sin(t)+\ln|\csc(t)|$$

and so:

$$I=\lim_{t\to0}\left(1-\sin(t)+\ln|\csc(t)| \right)=\infty$$

I left off the original coefficient of 4, as this did not impact the problem.