Tacx Trainer Software 4 0.rar
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Tacx Trainer Software 4 0.rar
Tacx Trainer Software 4 0.rar
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Tacx Trainer Software 4 0.rar
Tacx Trainer Software 4 0.rar
Tacx Trainer Software 4 0.rar
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Why is $\frac{1}{\pi}\int_{ -1}^1\frac{x^2}{1+x^2}dx$ different from $\frac{1}{\pi}\int_{0}^1\frac{x^2}{1+x^2}dx$?
I’m trying to evaluate $$I=\frac{1}{\pi}\int_{ -1}^1\frac{x^2}{1+x^2}dx$$ without using any complex analysis.
I’ve rewritten the integrand as follows:
$$\frac{x^2}{1+x^2}=\frac{x^2}{2}-\frac{1}{2}\cdot\frac{1}{1+x^2}=\frac{x^2}{2}-\frac{1}{2}\cdot\frac{1-x^2}{1-x^2}$$
Integrating by parts:
$$I=\frac{1}{2\pi}\int_{ -1}^1\frac{x^2}{2}\,dx+\frac{1}{2\pi}\int_{ -1}^1\frac{1}{1+x^2}\,dx-\frac{1}{\pi}\int_{ -1}^1\frac{1}{2}\cdot\frac{1-x^2}{1-x^2}dx$$
Intuitively I think the first integral must be 0, the second integral must be 1/4, but the last integral,
$$\begin{align}\left|\frac{1}{\pi}\int_{ -1}^1\frac{1}{1+x^2}\,dx\right|&\le\frac{1}{\pi}\int_{ -1}^1\frac{1}{1+x^2}dx\le\frac{1}{\pi}\int_{0}^1\frac{1}{1+x^2}dx\\&=\frac{1}{\pi}\cdot\frac{1}{2}\
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