พิสูจน์หน่อย

พิสูจน์ว่า อินทิเกรตของ (e^-3t* -e^-6t)/ t dt (t = 0 to infinity) = ln 2
thank you

[Anton]

Problem. Show that $$\int_0^\infty\,\frac{\mathrm{e}^{-3t}-\mathrm{e}^{-6t}}{t}\,\mathrm{d}t=\ln(2)\,.$$

Consider $f(x):=\displaystyle\int_0^\infty\,\dfrac{\exp(-t)-\exp(-xt)}{t}\,\mathrm{d}t$, where $x> 0$. Observe that
$$f'(x)=\int_0^\infty\,\frac{\partial}{\partial x}\,\left(\dfrac{\exp(-t)-\exp(-xt)}{t}\right)\,\mathrm{d}t=\int_0^\infty\,\exp(-xt)\,\mathrm{d}t=\frac{1}{x}\,.$$
Consequently,
$$f(x)-f(1)=\int_1^x\,\frac{1}{y}\,\mathrm{d}y=\ln(x)\,.$$
Because $f(1)=0$, we conclude that $f(x)=\ln(x)$ for every $x>0$. You ask for $$\int_0^\infty\,\frac{\exp(-3t)-\exp(-6t)}{t}\,\mathrm{d}t=f(6)-f(3)=\ln(6)-\ln(3)=\ln(2)\,.$$