integral

simplependulum

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Oh , i haven't used this method in solving integrals for a long time ..

Let \(\displaystyle I(b) = \int_0^{\infty} \frac{\ln(bx^6+1)}{x^6+1}~dx \)

The integral is equal to \(\displaystyle I(1) \) or \(\displaystyle I(1) - I(0) \)


Consider the derivative

\(\displaystyle I'(b) = \int_0^{\infty} \frac{x^6}{(x^6+1)(bx^6+1)}~dx \)

\(\displaystyle = \frac{1}{b-1} \int_0^{\infty} \left( \frac{1}{x^6+1} - \frac{1}{bx^6+1} \right) ~dx \)

Sub. \(\displaystyle \sqrt[6]{b} x = t \) on the second integral , we have

\(\displaystyle I'(b) = \frac{1}{b-1} \left( 1- \frac{1}{\sqrt[6]{b}} \right) \int_0^{\infty} \frac{dx}{x^6+1} \)


It is easy to check that \(\displaystyle \int_0^{\infty} \frac{dx}{x^6+1} = \frac{\pi}{3} \)

Therefore , \(\displaystyle I'(b) = \frac{\pi}{3} \frac{1}{b-1} \left( 1- \frac{1}{\sqrt[6]{b}} \right) \)

\(\displaystyle I = I(1) - I(0) = \int_0^1 \frac{\pi}{3}\frac{1}{b-1} \left( 1- \frac{1}{\sqrt[6]{b}} \right) db\)

Sub. \(\displaystyle b = t^6 ~,~ db = 6t^5~dt \)

\(\displaystyle I = 2\pi \int_0^1 \frac{t^5}{t^6-1} \left( 1 - \frac{1}{t} \right)~dt\)

\(\displaystyle = 2\pi \int_0^1 \frac{t^4(t-1)}{t^6-1}~dt\)

\(\displaystyle = \frac{\pi}{3} \int_0^1 \left[ \frac{2}{t+1} + \frac{4t^3 + 2t^2}{t^4 + t^2 + 1 } - \frac{ 3(t^2+1)+(t^2-1)}{t^4+t^2 + 1 } \right] ~dt \)

\(\displaystyle = \frac{\pi}{3} \left[ 2\ln(t+1) + \ln(t^4+t^2+1) - \sqrt{3} \tan^{-1}( \frac{t^2-1}{\sqrt{3} t } ) - \frac{1}{2} \ln \big{|} \frac{t^2-t +1}{t^2+t+1} \big{|} \right]_0^1 \)

\(\displaystyle = \frac{\pi}{3} [2 \ln{2} + \ln{3} - \frac{\sqrt{3}\pi}{2} + \frac{1}{2} \ln{3} ] \)

\(\displaystyle = \frac{\pi}{6} [ \ln{432} - \sqrt{3} \pi ] \)

ps

To show \(\displaystyle I = \int_0^{\infty} \frac{dx}{x^6+1} = \frac{\pi}{3} \)

sub. \(\displaystyle x = 1/t \) we have

\(\displaystyle I = \int_0^{\infty} \frac{t^4}{t^6+1}~dt \)

Then \(\displaystyle I - \int_0^{\infty} \frac{t^2}{t^6+1}~dt + I = \int_0^{\infty} \frac{t^4 - t^2 + 1}{t^6+1}~dt = \int_0^{\infty} \frac{dt}{t^2+1} = \frac{\pi}{2} \)

\(\displaystyle 2I = \frac{\pi}{2} + [ \frac{1}{3} \tan^{-1}(t^3) ]_0^{\infty} = \frac{2}{3}\pi \)

\(\displaystyle I = \frac{\pi}{3} \)
 
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