Evaluate $$ \int_0^\infty\frac{dx}{x^2-2x+4}. $$ I cannot figure it out. Any hint is appreciated.
Is it possible to calculate this integral using complex analysis?
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$\begingroup$
complex-analysis
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3The integral is divergent so it doesn't exist, no matter how you try to calculate it. – 2012-12-05
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0Sorry, I corrected the problem. I think I can do it now. – 2012-12-05
2 Answers
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No need for complex integration:
$$x^2-2x+4=(x-1)^2+3=3\left(1+\left(\frac{x-1}{\sqrt 3}\right)^2\right)\Longrightarrow$$
$$\int_0^\infty\frac{dx}{x^2-2x+4}=\frac{1}{\sqrt 3}\int_0^\infty\frac{\frac{1}{\sqrt 3}dx}{\left(1+\left(\frac{x-1}{\sqrt 3}\right)^2\right)}=$$
$$=\left.\frac{1}{\sqrt 3}\arctan\frac{x-1}{\sqrt 3}\right|_0^\infty=\frac{1}{\sqrt 3}\left(\frac{\pi}{2}-\left(-\frac{\pi}{6}\right)\right)=\frac{2\pi}{3\sqrt 3}$$
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0Thank you so much, @DonAntonio. – 2012-12-05
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Just in a nicer form: $$\int_{0}^{+\infty}\frac{dx}{x^2-2x+4}=\int_{-1}^{+\infty}\frac{dx}{x^2+3}=\frac{1}{\sqrt{3}}\int_{-1/\sqrt{3}}^{+\infty}\frac{dz}{z^2+1}=\frac{1}{\sqrt{3}}\left(\frac{\pi}{2}+\arctan\frac{1}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}\left(\pi-\arctan\sqrt{3}\right)=\frac{2\pi}{3\sqrt{3}}.$$
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0Thank you, Jack, your solution is also very good and innovative. – 2012-12-07