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

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

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TopNewestyou need knowledge of euler's formula and polylogarithm. first we write \[\tan(y)=x\to x=\arctan(x)=y\] or \[x=\dfrac{\sin(y)}{\cos(y)}=-i\dfrac{e^{2iy}-1}{e^{2iy}+1}\] So \[y=\dfrac{i}{2} \ln(1+ix)-\dfrac{i}{2} \ln(1-ix)\] so, \[\arctan(x)=\dfrac{i}{2} \ln(1+ix)-\dfrac{i}{2} \ln(1-ix)\] putting this in the integral \[I=\dfrac{i}{2}\int \dfrac{ \ln(1+ix)-\ln(1-ix)}{x}dx= \dfrac{i}{2}(li_2(-ix)-li_2(ix))+C\] – Aareyan Manzoor · 1 year ago

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– Pratik Roy · 1 year ago

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

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 · 1 year ago

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– Pratik Roy · 1 year ago

thanksLog in to reply

– Pratik Roy · 1 year ago

how can i calculate the integral if the limits are given?Log in to reply

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– Pratik Roy · 1 year ago

BUT THE LIMITS OF THIS SUM WAS FROM 1/2014 TO 2014Log in to reply

PS dont use all caps it is considered rude. – Aareyan Manzoor · 1 year ago

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– Aareyan Manzoor · 1 year ago

Ok, here it is: let \[I=\int_{1/2014}^{2014} \dfrac{\arctan(x)}{x} \, dx\] Use the sub u=1/x to get \[I=\int_{1/2014}^{2014} \dfrac{arccot(x)}{x} \, dx\] add these to get \[2I=\int_{1/2014}^{2014} \dfrac{\arctan(x)+arccot(x)}{x} \, dx\] Now apply integration by parts: \[\int f'g=fg-\int fg'\] \[f'=\dfrac{1}{x}\to f=\ln(x), g= \arctan(x)+arccot(x)\to g'=0\] now the integral is just \[2I=(\arctan(x)+arccot(x))ln(x)|_{1/2014}^{2014}-\int 0 dx=(\arctan(x)+arccot(x))ln(x)|_{1/2014}^{2014}\] I think you can put the limits yourself now.Log in to reply

\( tan^{-1}(x) + cot^{-1}(x) = \dfrac{\pi}{2} \) – Vighnesh Shenoy · 1 year ago

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– Aareyan Manzoor · 1 year ago

I had forgotten that formula :pLog in to reply

– Pratik Roy · 1 year ago

thanks a lotLog in to reply

– Pratik Roy · 1 year ago

can you also please tell me the solution of the another note i have posted named InequalityLog in to reply

– Pratik Roy · 1 year ago

can you please tell how to solve the inequality i have postedLog in to reply

– Aareyan Manzoor · 1 year ago

i will look into it.Log in to reply

– Pratik Roy · 1 year ago

thanksLog in to reply