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I was looking at this very famous answer and unfortunately I could not get through even the first step:

\begin{align} & 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_1^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\ &= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_0^{1} dt \frac{t^{1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\ \end{align}

Could someone please walk me through this step? how are we using the simmetry of $1/t$ here?

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Monolite
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1 Answers1

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In the integral

$$\int \limits _1 ^\infty \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \Bbb dt$$

make the change $u = \frac 1 t$. This will turn it into

$$\int \limits _1 ^0 \frac {u^{\frac 1 2}} {1 - \frac 1 {u^2}} \log \left( \frac {5 - \frac 2 u + \frac 1 {u^2}} {1 - \frac 2 u + \frac 5 {u^2}} \right) \left( - \frac 1 {u^2} \right) \Bbb d u = \int \limits _0 ^1 \frac {u^{\frac 1 2}} {1 - \frac 1 {u^2}} \log \left( \frac {5 - \frac 2 u + \frac 1 {u^2}} {1 - \frac 2 u + \frac 5 {u^2}} \right) \left( \frac 1 {u^2} \right) \Bbb d u = \\ \int \limits _0 ^1 \frac {u^{\frac 1 2}} {u^2 - 1} \log \left( \frac {5u^2 - 2u + 1} {u^2 - 2u + 5} \right) \Bbb d u = - \int \limits _0 ^1 \frac {u^{\frac 1 2}} {u^2 - 1} \log \left( \frac {u^2 - 2u + 5} {5u^2 - 2u + 1} \right) \Bbb d u = \\ \int \limits _0 ^1 \frac {u^{\frac 1 2}} {1 - u^2} \log \left( \frac {5 - 2u + u^2} {1 - 2u + 5u^2} \right) \Bbb d u .$$

Now change the letter $u$ into $t$ and you get the second integral on the second row of your question.

Alex M.
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  • thank you greatly, what is the symmetry we are using? I mean is there a geometric reason that most of the integrand function remains the same? – Monolite Mar 17 '17 at 14:53
  • @Monolite: I don't see any deep meaning begind this; the integrand is purposefully conceived to have this symmetry, and thus make that problem solvable. Nothing more, in my opinion. – Alex M. Mar 17 '17 at 15:05
  • Thanks, it made me wonder because a bit further down solving the same integral he cites again the simmetry of the map $z \rightarrow 1/z$ to quickly prove that an integral is zero. So even in that case he is simply making a u substitution correct? – Monolite Mar 17 '17 at 15:09