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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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"Am I breaking the integral up into four integrations..." - yep, you interpreted correctly. Try it out! –  J. M. Feb 18 '12 at 3:53
    
For the first question: what values of $t$ will make $\dfrac{t}{1-t^2}$ take the values $-\infty$ and $+\infty$? –  J. M. Feb 18 '12 at 3:56
    
@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). –  Leslie Feb 18 '12 at 13:35
    
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... :) –  J. M. Feb 18 '12 at 13:38
    
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). –  Leslie Feb 18 '12 at 13:45

1 Answer 1

$\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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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... –  J. M. Feb 18 '12 at 15:03
    
I realized after I posted which means that this integrable is not solvable using trapezoidal approximation using $x_0 = a$ and $x_n=b$. –  Leslie Feb 18 '12 at 18:15
    
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. –  J. M. Feb 18 '12 at 23:07
    
You are correct, something went over my head. –  Leslie Feb 18 '12 at 23:43
    
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. –  J. M. Feb 18 '12 at 23:57

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