Can anyone help to integrat this function please?ic∫−11(−2x11+(tcx)2)dtC&gt;0, x&gt;0 and i the imaginary partNot really sure how to go about integrating it.

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=tcx , as suggested by Gregor Bruns, to get:ic∫−11−2x11+(tcx)2=−2icx∫−cxcx11+u2⋅xcdu=−2i∫−cxcx11+u2duNow you should recognise the integrand as ddutan−1(u), so we have−2i[tan−1(cx)−tan−1(−cx)]=−4itan−1(cx)

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eketsaod7

Answered question

2022-03-18

Can anyone help to integrat this function please?

i c \int_{- 1}^{1} (\frac{- 2}{x} \frac{1}{1 + {(\frac{t c}{x})}^{2}}) d t

C>0, x>0 and i the imaginary part
Not really sure how to go about integrating it.

Answer & Explanation

Brock Floyd

Beginner2022-03-19Added 6 answers

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{t c}{x}

, as suggested by Gregor Bruns, to get:

i c \int_{- 1}^{1} \frac{- 2}{x} \frac{1}{1 + {(\frac{t c}{x})}^{2}} = \frac{- 2 i c}{x} \int_{- \frac{c}{x}}^{\frac{c}{x}} \frac{1}{1 + u^{2}} \cdot \frac{x}{c} d u = - 2 i \int_{- \frac{c}{x}}^{\frac{c}{x}} \frac{1}{1 + u^{2}} d u

Now you should recognise the integrand as

\frac{d}{d u} \tan^{- 1} (u)

, so we have

- 2 i [\tan^{- 1} (\frac{c}{x}) - \tan^{- 1} (- \frac{c}{x})] = - 4 i \tan^{- 1} (\frac{c}{x})

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