Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
"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

$\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)] $

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.