Murder the Beast

Calculus (Integrals, Series)
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Tolaso J Kos
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Murder the Beast

#1

Post by Tolaso J Kos »

Someone to murder this "beast"?

$$ \displaystyle \mathcal{J}=\int_{0}^{\pi/4}\left [ \frac{\left ( 1-x^2 \right )\ln\left ( 1+x^2 \right )+\left ( 1+x^2 \right )-\left ( 1-x^2 \right )\ln\left ( 1-x^2 \right )}{\left ( 1-x^4 \right )\left ( 1+x^2 \right )} \right ]x\, {\rm exp} \left [ \frac{x^2-1}{x^2+1} \right ]\, dx $$
Imagination is much more important than knowledge.
Papapetros Vaggelis
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Re: Murder the Beast

#2

Post by Papapetros Vaggelis »

The function \(\displaystyle{f:\left[0,\frac{\pi}{4}\right]\longrightarrow \mathbb{R}}\) defined by

\(\displaystyle{f(x)=\left[\dfrac{\left(1-x^2\right)\,\ln\,\left(1+x^2\right)+\left(1+x^2\right)-\left(1-x^2\right)\,\ln\,\left(1-x^2\right)}{\left(1-x^4\right)\,\left(1+x^2\right)}\right]\,x\,\exp\,\left[\dfrac{x^2-1}{x^2+1}\right]}\)

is continuous at \(\displaystyle{\left[0,\frac{\pi}{4}\right]}\) , so \(\displaystyle{J<\infty}\) .

Also,

\(\displaystyle{\dfrac{\rm{d}}{\rm{dx}}\left[\ln\,\left(1+x^2\right)-\ln\,\left(1-x^2\right)\right]=\dfrac{2\,x}{1+x^2}+\dfrac{2\,x}{1-x^2}=\dfrac{4\,x}{\left(1+x^2\right)\,\left(1-x^2\right)}}\)

and

\(\displaystyle{\dfrac{\rm{d}}{\rm{dx}}\,\left[\dfrac{x^2-1}{x^2+1}\right]=\dfrac{\rm{d}}{\rm{dx}}\,\left[1-\dfrac{2}{x^2+1}\right]=\dfrac{4\,x}{\left(x^2+1\right)^2}}\)

Therefore,

\(\displaystyle{\begin{aligned} J&=\int_{0}^{\frac{\pi}{4}}\left[\dfrac{\left(1-x^2\right)\,\ln\,\left(1+x^2\right)+\left(1+x^2\right)-\left(1-x^2\right)\,\ln\,\left(1-x^2\right)}{\left(1-x^4\right)\,\left(1+x^2\right)}\right]\,x\,\exp\,\left[\dfrac{x^2-1}{x^2+1}\right]\,\rm{dx}\\&=\int_{0}^{\frac{\pi}{4}}\left[\dfrac{\left(1-x^2\right)\,\left(\ln\,\left(1+x^2\right)-\ln\,\left(1-x^2\right)\right)+\left(1+x^2\right)}{\left(1-x^2\right)\,\left(1+x^2\right)^2}\right]\,x\,\exp\,\left[\dfrac{x^2-1}{x^2+1}\right]\,\rm{dx}\\&=\frac{1}{4}\,\int_{0}^{\frac{\pi}{4}}\left[\dfrac{4\,x}{\left(1+x^2\right)^2}\,\exp\,\left(\dfrac{x^2-1}{x^2+1}\right)\,\left(\ln\,\left(1+x^2\right)-\ln\,\left(1-x^2\right)\right)+\dfrac{4\,x}{\left(1-x^2\right)\,\left(1+x^2\right)}\,\exp\,\left(\dfrac{x^2-1}{x^2+1}\right)\right]\,\rm{dx}\\&=\frac{1}{4}\,\int_{0}^{\frac{\pi}{4}}\left[\left(\dfrac{x^2-1}{x^2+1}\right)'\,\exp\,\left(\dfrac{x^2-1}{x^2+1}\right)\,\left(\ln\,\left(1+x^2\right)-\ln\,\left(1-x^2\right)\right)+\exp\,\left(\dfrac{x^2-1}{x^2+1}\right)\,\left(\ln\,\left(\dfrac{1+x^2}{1-x^2}\right)\right)'\right]\,\rm{dx}\\&=\frac{1}{4}\,\int_{0}^{\frac{\pi}{4}}\left[\exp\,\left(\dfrac{x^2-1}{x^2+1}\right)\,\ln\,\left(\dfrac{1+x^2}{1-x^2}\right)\right]'\,\rm{dx}\\&=\left[\frac{1}{4}\,\exp\,\left(\dfrac{x^2-1}{x^2+1}\right)\,\ln\,\left(\dfrac{1+x^2}{1-x^2}\right)\right]_{0}^{\frac{\pi}{4}}\\&=\frac{1}{4}\,\exp\,\left(\dfrac{\pi^2-16}{\pi^2+16}\right)\,\ln\,\left(\dfrac{16+\pi^2}{16-\pi^2}\right)\end{aligned}}\)
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