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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.

Thanks

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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

1 Answer 1

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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 $$-2i\left[\tan^{-1}(c/x)-\tan^{-1}(-c/x)\right]=-4i\tan^{-1}(c/x)$$

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Thank you for your help. –  CJC Dec 2 '12 at 17:02

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