Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Evaluate $$ \int_0^\infty\frac{dx}{x^2-2x+4}. $$ I cannot figure it out. Any hint is appreciated.

share|cite|improve this question
The integral is divergent so it doesn't exist, no matter how you try to calculate it. – DonAntonio Dec 5 '12 at 23:01
Sorry, I corrected the problem. I think I can do it now. – Sam Dec 5 '12 at 23:06

2 Answers 2

up vote 4 down vote accepted

No need for complex integration:

$$x^2-2x+4=(x-1)^2+3=3\left(1+\left(\frac{x-1}{\sqrt 3}\right)^2\right)\Longrightarrow$$

$$\int_0^\infty\frac{dx}{x^2-2x+4}=\frac{1}{\sqrt 3}\int_0^\infty\frac{\frac{1}{\sqrt 3}dx}{\left(1+\left(\frac{x-1}{\sqrt 3}\right)^2\right)}=$$

$$=\left.\frac{1}{\sqrt 3}\arctan\frac{x-1}{\sqrt 3}\right|_0^\infty=\frac{1}{\sqrt 3}\left(\frac{\pi}{2}-\left(-\frac{\pi}{6}\right)\right)=\frac{2\pi}{3\sqrt 3}$$

share|cite|improve this answer
Thank you so much, @DonAntonio. – Sam Dec 5 '12 at 23:20
Any time......... – DonAntonio Dec 5 '12 at 23:22

Just in a nicer form: $$\int_{0}^{+\infty}\frac{dx}{x^2-2x+4}=\int_{-1}^{+\infty}\frac{dx}{x^2+3}=\frac{1}{\sqrt{3}}\int_{-1/\sqrt{3}}^{+\infty}\frac{dz}{z^2+1}=\frac{1}{\sqrt{3}}\left(\frac{\pi}{2}+\arctan\frac{1}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}\left(\pi-\arctan\sqrt{3}\right)=\frac{2\pi}{3\sqrt{3}}.$$

share|cite|improve this answer
Thank you, Jack, your solution is also very good and innovative. – Sam Dec 7 '12 at 3:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.