$$\int_{0}^{\infty}\frac{dx}{(1+x^{2})(1+x^{r})}$$
$$=\int_{0}^{\frac{\pi}{2}}\frac{du}{1+tan^{r}u}$$
$$=\int_{0}^{\frac{\pi}{2}}\frac{cos^{r}u}{sin^{r}u+cos^{r}u}du$$
$$=\int_{0}^{\frac{\pi}{2}}\frac{sin^{r}u}{sin^{r}u+cos^{r}u}du$$
$$\therefore\int_{0}^{\infty}\frac{dx}{(1+x^{2})(1+x^{r})}=\frac{1}{2}\left[\int_{0}^{\frac{\pi}{2}}\frac{cos^{r}u}{sin^{r}u+cos^{r}u}du+\int_{0}^{\frac{\pi}{2}}\frac{sin^{r}u}{sin^{r}u+cos^{r}u}du\right]=\frac{\pi}{4}$$
__________________
$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x-b\sin x}{a\sin x+b\cos x}dx=\ln\left(\frac{a}{b}\right)$$
BUT $$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x+b\sin x}{a\sin x+b\cos x}dx=\frac{\pi ab}{a^{2}+b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\ln\left(\frac{a}{b}\right)$$
14 เมษายน 2008 11:39 : ข้อความนี้ถูกแก้ไขแล้ว 1 ครั้ง, ครั้งล่าสุดโดยคุณ Timestopper_STG
|