Convergence (Evaluation)
- Tolaso J Kos
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Convergence (Evaluation)
Examine whether the following integral $$\int_{0}^{\infty}\frac{dx}{x^4-x^3+x^2-x}$$ converges or not. In case it does, evaluate it.
Imagination is much more important than knowledge.
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Re: Convergence (Evaluation)
For each \(\displaystyle{x\in\mathbb{R}}\) holds :
\(\displaystyle{f(x)=x^4-x^3+x^2-x=x^3\,(x-1)+x\,(x-1)=x\,(x-1)\,(x^2+1)}\), so we have :
\(\displaystyle{\int_{0}^{\infty}\dfrac{1}{x^4-x^3+x^2-x}\,\mathrm{d}x=\int_{0}^{1/2}\dfrac{1}{f(x)}\,\mathrm{d}x+\int_{1/2}^{1}\dfrac{1}{f(x)}\,\mathrm{d}x+\int_{1}^{2}\dfrac{1}{f(x)}\,\mathrm{d}x+\int_{2}^{\infty}\dfrac{1}{f(x)}\,\mathrm{d}x}\)
Let \(\displaystyle{a\,,b\,,c\,,d\in\mathbb{R}}\) such that :
\(\displaystyle{\dfrac{1}{f(x)}=\dfrac{a}{x}+\dfrac{b}{x-1}+\dfrac{c\,x+d}{x^2+1}\,\,\forall\,x\in\mathbb{R}-\left\{0,1\right\}}\) .
It's easy to see that \(\displaystyle{\left(a,b,c,d\right)=\left(-1,\dfrac{1}{2},\dfrac{1}{2},-\dfrac{1}{2}\right)}\) and then :
if \(\displaystyle{a\,,b\in\mathbb{R}\cap\left(0,+\infty\right)\,,a<b}\) :
\(\displaystyle{\begin{aligned}\int_{a}^{b}\dfrac{1}{f(x)}\,\mathrm{d}x&=\int_{a}^{b}\,\left[-\dfrac{1}{x}+\dfrac{1}{2\,(x-1)}+\dfrac{x-1}{2\,(x^2+1)}\right)\,\mathrm{d}x\\&=\int_{a}^{b}\,\left[-\dfrac{1}{x}+\dfrac{1}{2\,(x-1)}+\dfrac{2\,x}{4\,(x^2+1)}-\dfrac{1}{2\,(x^2+1}\right)\,\mathrm{d}x\\&=\left[-\ln\,x+\dfrac{1}{2}\,\ln\,(x-1)+\dfrac{1}{4}\,\ln\,(x^2+1)-\dfrac{1}{2}\,\arctan\,x\right]_{a}^{b}\\&=\left[\dfrac{1}{2}\,\ln\,\dfrac{x-1}{x^2}+\dfrac{1}{4}\,\ln\,(x^2+1)-\dfrac{1}{2}\,\arctan\,x\right]_{a}^{b}\\&=\left[\dfrac{1}{4}\,\ln\,\dfrac{(x^2+1)\,(x-1)^2}{x^4}-\dfrac{1}{2}\,\arctan\,x\right]_{a}^{b}\end{aligned}}\)
Now, we get :
\(\displaystyle{\int_{0}^{1/2}\dfrac{1}{f(x)}\,\mathrm{d}x\to-\infty,\int_{1/2}^{1}\dfrac{1}{f(x)}\,\mathrm{d}x\to -\infty\,\,,\int_{1}^{2}\dfrac{1}{f(x)}\,\mathrm{d}x\to +\infty\,\,,\int_{2}^{+\infty}\dfrac{1}{f(x)}\,\mathrm{d}x<\infty}\)
So, what can we deduce about the initial integral ?
\(\displaystyle{f(x)=x^4-x^3+x^2-x=x^3\,(x-1)+x\,(x-1)=x\,(x-1)\,(x^2+1)}\), so we have :
\(\displaystyle{\int_{0}^{\infty}\dfrac{1}{x^4-x^3+x^2-x}\,\mathrm{d}x=\int_{0}^{1/2}\dfrac{1}{f(x)}\,\mathrm{d}x+\int_{1/2}^{1}\dfrac{1}{f(x)}\,\mathrm{d}x+\int_{1}^{2}\dfrac{1}{f(x)}\,\mathrm{d}x+\int_{2}^{\infty}\dfrac{1}{f(x)}\,\mathrm{d}x}\)
Let \(\displaystyle{a\,,b\,,c\,,d\in\mathbb{R}}\) such that :
\(\displaystyle{\dfrac{1}{f(x)}=\dfrac{a}{x}+\dfrac{b}{x-1}+\dfrac{c\,x+d}{x^2+1}\,\,\forall\,x\in\mathbb{R}-\left\{0,1\right\}}\) .
It's easy to see that \(\displaystyle{\left(a,b,c,d\right)=\left(-1,\dfrac{1}{2},\dfrac{1}{2},-\dfrac{1}{2}\right)}\) and then :
if \(\displaystyle{a\,,b\in\mathbb{R}\cap\left(0,+\infty\right)\,,a<b}\) :
\(\displaystyle{\begin{aligned}\int_{a}^{b}\dfrac{1}{f(x)}\,\mathrm{d}x&=\int_{a}^{b}\,\left[-\dfrac{1}{x}+\dfrac{1}{2\,(x-1)}+\dfrac{x-1}{2\,(x^2+1)}\right)\,\mathrm{d}x\\&=\int_{a}^{b}\,\left[-\dfrac{1}{x}+\dfrac{1}{2\,(x-1)}+\dfrac{2\,x}{4\,(x^2+1)}-\dfrac{1}{2\,(x^2+1}\right)\,\mathrm{d}x\\&=\left[-\ln\,x+\dfrac{1}{2}\,\ln\,(x-1)+\dfrac{1}{4}\,\ln\,(x^2+1)-\dfrac{1}{2}\,\arctan\,x\right]_{a}^{b}\\&=\left[\dfrac{1}{2}\,\ln\,\dfrac{x-1}{x^2}+\dfrac{1}{4}\,\ln\,(x^2+1)-\dfrac{1}{2}\,\arctan\,x\right]_{a}^{b}\\&=\left[\dfrac{1}{4}\,\ln\,\dfrac{(x^2+1)\,(x-1)^2}{x^4}-\dfrac{1}{2}\,\arctan\,x\right]_{a}^{b}\end{aligned}}\)
Now, we get :
\(\displaystyle{\int_{0}^{1/2}\dfrac{1}{f(x)}\,\mathrm{d}x\to-\infty,\int_{1/2}^{1}\dfrac{1}{f(x)}\,\mathrm{d}x\to -\infty\,\,,\int_{1}^{2}\dfrac{1}{f(x)}\,\mathrm{d}x\to +\infty\,\,,\int_{2}^{+\infty}\dfrac{1}{f(x)}\,\mathrm{d}x<\infty}\)
So, what can we deduce about the initial integral ?
- Grigorios Kostakos
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Re: Convergence (Evaluation)
Even one of the four integrals \[\displaystyle\int_{0}^{1/2}\dfrac{1}{f(x)}\,\mathrm{d}x,\quad\int_{1/2}^{1}\dfrac{1}{f(x)}\,\mathrm{d}x\,\,,\quad \int_{1}^{2}\dfrac{1}{f(x)}\,\mathrm{d}x\,\,,\quad\int_{2}^{+\infty}\dfrac{1}{f(x)}\,\mathrm{d}x \] diverges, then the integral \[\displaystyle\int_{0}^{+\infty}\dfrac{1}{f(x)}\,\mathrm{d}x\] diverges.Papapetros Vaggelis wrote:..So, what can we deduce about the initial integral ?
So, the answer is obvious.
Grigorios Kostakos
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