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So I have a two fold question, one I believe is simple but my algebra seems to be off, the other involves the trapezoidal rule of integration using Mathematica as an aid. Here they are:

$1.\quad \displaystyle \int_{-\infty}^{\infty} \frac{\operatorname{sech}(x)}{x^2+1} dx = \int_{-1}^{1} \operatorname{sech}\left(\frac{1}{1-t^2}\right)\frac{t^2+1}{t^4-t^2+1} dt$

I know I need to let $x = \frac{t}{1-t^2}$ and take the limits as $t \to \infty$,change the limits of integration and do the same for $t \to -\infty$ but I can't seem to nail it down. Why are my limits going to be $-1$ and $1$?

$2$. Space five points equally from $-1$ to $1$ and compute the four trapezoid approximation of $\int_{-1}^{1} \mathrm{sech}(\frac{1}{1-t^2})\frac{t^2+1}{t^4-t^2+1} dt$ using Mathematica to evaluate $\operatorname{sech}(x)$. To be honest, I'm not really sure what the question is asking. Am I breaking the integral up into four integrations the first of which is from $-1$ to $-0.5$? How do I use Mathematica to evaluate? Any help is appreciated.

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  • $\begingroup$ "Am I breaking the integral up into four integrations..." - yep, you interpreted correctly. Try it out! $\endgroup$ – J. M. isn't a mathematician Feb 18 '12 at 3:53
  • $\begingroup$ For the first question: what values of $t$ will make $\dfrac{t}{1-t^2}$ take the values $-\infty$ and $+\infty$? $\endgroup$ – J. M. isn't a mathematician Feb 18 '12 at 3:56
  • $\begingroup$ @J.M. Thank you. I entered the integral from $-1$ to $-0.5$ into Mathematica but it gives me back an answer which still involves sech (using the original function). $\endgroup$ – Leslie Feb 18 '12 at 13:35
  • $\begingroup$ the second question was asking you to use the trapezoidal rule over the four separate panels you made out of the interval $(-1,1)$ for the evaluation, and not an analytical evaluation... :) $\endgroup$ – J. M. isn't a mathematician Feb 18 '12 at 13:38
  • $\begingroup$ LOL. I didn't clarify. This is for one of my tutoring students and the professor specifically requests that the students use Wolfram Alpha or Mathematica to evaluate sech(x). $\endgroup$ – Leslie Feb 18 '12 at 13:45
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$\quad \displaystyle \int_{-1}^{1} \mathrm{sech}\left(\frac{1}{1-t^2}\right)\frac{t^2+1}{t^4-t^2+1} dt = \lim n \to \infty \frac{b-a}{2n}[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{(n-1)}) + f(x_n)] $

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  • $\begingroup$ Er, it's $\mathrm{sech}\left(\frac{1}{1-t^2}\right)\frac{t^2+1}{t^4-t^2+1}$ you're supposed to be evaluating at those five points... $\endgroup$ – J. M. isn't a mathematician Feb 18 '12 at 15:03
  • $\begingroup$ I realized after I posted which means that this integrable is not solvable using trapezoidal approximation using $x_0 = a$ and $x_n=b$. $\endgroup$ – Leslie Feb 18 '12 at 18:15
  • $\begingroup$ It still is. It looks as if you missed the point of my question if you know the limit of the hyperbolic secant as the argument increases without bound. $\endgroup$ – J. M. isn't a mathematician Feb 18 '12 at 23:07
  • $\begingroup$ Alright... so tell me, what is the limit of the hyperbolic secant? That's basically how to interpret "$\mathrm{sech}(\infty)$" in the trapezoidal approximation you have obtained. The other points can be numerically evaluated, as was asked in your question. $\endgroup$ – J. M. isn't a mathematician Feb 18 '12 at 23:57
  • $\begingroup$ The limit is zero. I think I'm just having trouble in general grasping the concept of the trapezoidal approximation when the endpoints are not continuous. Can I just take the limit from one side? $\endgroup$ – Leslie Feb 19 '12 at 0:12

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