Integration

arctan(x)xdx\large \int \dfrac{\arctan(x)}{x} \, dx Can anyone tell how this integration can be done?

Note by Pratik Roy
3 years, 2 months ago

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you need knowledge of euler's formula and polylogarithm. first we write tan(y)=xx=arctan(x)=y\tan(y)=x\to x=\arctan(x)=y or x=sin(y)cos(y)=ie2iy1e2iy+1x=\dfrac{\sin(y)}{\cos(y)}=-i\dfrac{e^{2iy}-1}{e^{2iy}+1} So y=i2ln(1+ix)i2ln(1ix)y=\dfrac{i}{2} \ln(1+ix)-\dfrac{i}{2} \ln(1-ix) so, arctan(x)=i2ln(1+ix)i2ln(1ix)\arctan(x)=\dfrac{i}{2} \ln(1+ix)-\dfrac{i}{2} \ln(1-ix) putting this in the integral I=i2ln(1+ix)ln(1ix)xdx=i2(li2(ix)li2(ix))+CI=\dfrac{i}{2}\int \dfrac{ \ln(1+ix)-\ln(1-ix)}{x}dx= \dfrac{i}{2}(li_2(-ix)-li_2(ix))+C

Aareyan Manzoor - 3 years, 2 months ago

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can you please suggest me from where i could learn more about doing these type of integrations.

Pratik Roy - 3 years, 2 months ago

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from here try the integration tricks wikis and advance integration. while this is not complete yet the contents there are good enough for this problem.

Aareyan Manzoor - 3 years, 2 months ago

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@Aareyan Manzoor thanks

Pratik Roy - 3 years, 2 months ago

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@Pratik Roy how can i calculate the integral if the limits are given?

Pratik Roy - 3 years, 2 months ago

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