อ้างอิง:
ข้อความเดิมเขียนโดยคุณ Timestopper_STG
74.จงหาค่าของ$\displaystyle{\int\limits_0^{\infty}\frac{e^{-t}-2e^{-3t}+e^{-5t}}{t^2}dt}$  |
At first we observe that $\displaystyle{\frac{1}{t^2}=\int_0^{\infty}xe^{-tx}dx}$
$\displaystyle{\int\limits_0^{\infty}\frac{e^{-t}-2e^{-3t}+e^{-5t}}{t^2}dt}$
$\displaystyle{=\int_0^{\infty}\int_0^{\infty}xe^{-tx}\left(e^{-t}-2e^{-3t}+e^{-5t}\right)dxdt}$
$\displaystyle{=\int_0^{\infty}x\int_0^{\infty}e^{-(1+x)t}-2e^{-(3+x)t}+e^{-(5+x)t}dtdx}$
$\displaystyle{=\int_0^{\infty}\left[\left(\frac{x}{1+x}\right)-2\left(\frac{x}{3+x}\right)+\left(\frac{x}{5+x}\right)\right]dx}$
$\displaystyle{=\int_0^{\infty}\left[\left(1-\frac{1}{1+x}\right)-2\left(1-\frac{3}{3+x}\right)+\left(1-\frac{5}{5+x}\right)\right]dx}$
$\displaystyle{=\int_0^{\infty}\left[-\frac{1}{1+x}+\frac{6}{3+x}-\frac{5}{5+x}dx\right]}$
$\displaystyle{=\left[\ln\left(\frac{(3+x)^6}{(1+x)(5+x)^5}\right)\right]_0^{\infty}=\ln\left(\frac{5^5}{3^6}\right)}$
Don't be surprise why a "ลมปราณบริสุทธิ์" man can solve this question with a beautiful method.
Because I have just copied this from the other place.
