Ngô Quốc Anh

September 23, 2007

Một bài thi OLP’ SV

Filed under: Giải Tích 2 — Ngô Quốc Anh @ 14:55

Chứng minh rằng \forall a\in\mathbb{R},


Chứng minh.

Using x=\tan (\theta )

\int_{0}^{\infty }\frac{1}{\left(x^{2}+1\right)\left(x^{\alpha }+1\right)}\, dx=\int_{0}^{\frac{\pi }{2}}\frac{1}{\tan^{\alpha }(\theta )+1}\, d\theta

P_{1}=\int_{0}^{\frac{\pi }{2}}\frac{1}{\tan^{\alpha }(\theta )+1}\, d\theta =\int_{0}^{\frac{\pi }{2}}\frac{\cos^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta

Putting \theta\to\frac{\pi }{2}-\theta

P_{2}=\int_{0}^{\frac{\pi }{2}}\frac{1}{\cot^{\alpha }(\theta )+1}\, d\theta =\int_{0}^{\frac{\pi }{2}}\frac{\sin^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta

Of course P_{1}=P_{2}

P_{1}+P_{2}=\int_{0}^{\frac{\pi }{2}}\frac{\cos^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta+\int_{0}^{\frac{\pi }{2}}\frac{\sin^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta =\int_{0}^{\frac{\pi }{2}}\frac{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta 
P_{1}+P_{2}=\int_{0}^{\frac{\pi }{2}}1\, d\theta =\frac{\pi }{2}=2 P_{1}=2 P_{2} 
And we are done.

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