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# Integration

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

Note by Pratik Roy
1 year ago

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you 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 $I=\dfrac{i}{2}\int \dfrac{ \ln(1+ix)-\ln(1-ix)}{x}dx= \dfrac{i}{2}(li_2(-ix)-li_2(ix))+C$ · 1 year ago

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. · 1 year ago

thanks · 1 year ago

how can i calculate the integral if the limits are given? · 1 year ago

Comment deleted Apr 06, 2016

BUT THE LIMITS OF THIS SUM WAS FROM 1/2014 TO 2014 · 1 year ago

ignore what i said, for those limits it is defined, wait as few minute as i write the solution

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

Comment deleted Apr 06, 2016

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. · 1 year ago

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

I had forgotten that formula :p · 1 year ago

thanks a lot · 1 year ago

can you also please tell me the solution of the another note i have posted named Inequality · 1 year ago

can you please tell how to solve the inequality i have posted · 1 year ago

i will look into it. · 1 year ago