# Ngô Quốc Anh

## September 23, 2007

### Một bài thi OLP’ SV

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

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

$\int_{0}^{\infty}\frac{dx}{(1+x^{2})(1+x^{a})}=\frac{\pi}{4}$.

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.