Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Can anyone help to integrat this function please?

$$ ic \int_{-1}^{1} \left(\frac{-2}{x} \frac{1} {1+(\frac{tc}{x})^2}\right)dt $$

C>0, x>0 and i the imaginary part

Not really sure how to go about integrating it.


share|cite|improve this question
Did you try a substitution $u = tc/x$? Afterwards, you can recognize the integrand as the derivative of $\arctan(u)$. – Gregor Bruns Nov 22 '12 at 18:12
up vote 0 down vote accepted

If $x$ is independent of t, and we have no reason to suppose it isn't, then we can factor it out of the integral and make the substitution $u=\frac{tc}{x}$, as suggested by Gregor Bruns, to get: $$ic\int_{-1}^{1}\frac{-2}{x} \frac{1}{1+\left(\frac{tc}{x}\right)^{2}}=\frac{-2ic}{x}\int_{-c/x}^{c/x}\frac{1}{1+u^{2}}\cdot\frac{x}{c}du=-2i\int_{-c/x}^{c/x}\frac{1}{1+u^{2}}du$$ Now you should recognise the integrand as $\frac{d}{du}\tan^{-1}(u)$, so we have $$-2i\left[\tan^{-1}(c/x)-\tan^{-1}(-c/x)\right]=-4i\tan^{-1}(c/x)$$

share|cite|improve this answer
Thank you for your help. – CJC Dec 2 '12 at 17:02

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.