# real value function (integral)

#### gomes

How would I do this? Im really stuck. Most appreciated!

#### zzzoak

$$\displaystyle \frac{df}{d \alpha} \: = \: \frac{1}{\alpha}$$
Integrating
$$\displaystyle f(\alpha) \: = \: ln | \alpha | +C$$
$$\displaystyle f(1) \: = \: C=0$$

$$\displaystyle f(\alpha \beta) \: = \: ln | \alpha | +ln | \beta |$$
$$\displaystyle f(\alpha \beta) \: = \: f(\alpha)+f(\beta)$$

#### Random Variable

If you wanted to evaluate it, you could write it as $$\displaystyle \int^{\infty}_{0} \int^{\alpha}_{1} e^{-tx} \ dt \ dx$$

then switch the order of integration

$$\displaystyle = \int^{\alpha}_{1} \int_{0}^{\infty} e^{-tx} \ dx \ dt$$

$$\displaystyle = \int^{\alpha}_{1} \frac{1}{t} \ dx = \ln \alpha$$

gomes

#### simplependulum

MHF Hall of Honor
No , the question says without evaluating the integral , we cannot do it in this way .

$$\displaystyle f( a) = \int_0^{\infty} \frac{ e^{-x} - e^{-ax} }{x}~dx$$

Sub. $$\displaystyle x = bt$$ and change the dummy variable , we have

$$\displaystyle f(a) = \int_0^{\infty} \frac{ e^{-bx} - e^{-abx} }{bx}~(bdx)$$

$$\displaystyle = \int_0^{\infty} \frac{ e^{-bx} - e^{-abx} }{x}~dx$$

$$\displaystyle = \int_0^{\infty} \frac{ e^{-bx} - e^{-x} + e^{-x} - e^{-abx} }{x}~dx$$

$$\displaystyle =\int_0^{\infty} \frac{ (e^{-x} - e^{-abx})-(e^{-x} - e^{-bx} ) }{x}~dx$$

$$\displaystyle = f(ab) - f(b)$$ so $$\displaystyle f(ab) = f(a) + f(b)$$