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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
8 months, 1 week 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 \[I=\dfrac{i}{2}\int \dfrac{ \ln(1+ix)-\ln(1-ix)}{x}dx= \dfrac{i}{2}(li_2(-ix)-li_2(ix))+C\] Aareyan Manzoor · 8 months, 1 week ago

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@Aareyan Manzoor can you please suggest me from where i could learn more about doing these type of integrations. Pratik Roy · 8 months, 1 week ago

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@Pratik Roy 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 · 8 months, 1 week ago

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@Aareyan Manzoor thanks Pratik Roy · 8 months, 1 week ago

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@Pratik Roy how can i calculate the integral if the limits are given? Pratik Roy · 8 months ago

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Comment deleted 8 months ago

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@Aareyan Manzoor BUT THE LIMITS OF THIS SUM WAS FROM 1/2014 TO 2014 Pratik Roy · 8 months ago

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@Pratik Roy 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. Aareyan Manzoor · 8 months ago

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@Pratik Roy 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. Aareyan Manzoor · 8 months ago

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@Aareyan Manzoor thanks a lot Pratik Roy · 8 months ago

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@Aareyan Manzoor Instead of applying by parts,
\( tan^{-1}(x) + cot^{-1}(x) = \dfrac{\pi}{2} \) Vighnesh Shenoy · 8 months ago

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@Vighnesh Shenoy I had forgotten that formula :p Aareyan Manzoor · 8 months ago

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@Pratik Roy can you also please tell me the solution of the another note i have posted named Inequality Pratik Roy · 8 months ago

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@Pratik Roy can you please tell how to solve the inequality i have posted Pratik Roy · 8 months ago

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@Pratik Roy i will look into it. Aareyan Manzoor · 8 months ago

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@Aareyan Manzoor thanks Pratik Roy · 8 months ago

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