Definite Integral Identity

Posted on February 6, 2018 by Sumant Sumant

Prove that \int_{0}^{\infty}\dfrac{1}{1+x^2}dx = 2\int_{0}^{1}\dfrac{1}{1+x^2}dx.
\Rightarrow \int_{0}^{\infty}\dfrac{1}{1+x^2}dx
\Rightarrow \int_{0}^{1}\dfrac{1}{1+x^2}dx+\int_{1}^{\infty}\dfrac{1}{1+x^2}dx
\Rightarrow \int_{0}^{1}\dfrac{1}{1+x^2}dx+\int_{1}^{0}\dfrac{1}{1+\frac{1}{t^2}}\frac{-1}{t^2}
\Rightarrow \int_{0}^{1}\dfrac{1}{1+x^2}dx+\int_{0}^{1}\dfrac{1}{1+t^2}dt
\Rightarrow \int_{0}^{1}\dfrac{1}{1+x^2}dx+\int_{0}^{1}\dfrac{1}{1+x^2}dx
\Rightarrow 2\int_{0}^{1}\dfrac{1}{1+x^2}dx

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