# Thread: real value function (integral)

1. ## real value function (integral)

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

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

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

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

then switch the order of integration

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

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

4. thanks! If possible, could you help me in this thread? Thanks

http://www.mathhelpforum.com/math-he...tive-help.html

5. No , the question says without evaluating the integral , we cannot do it in this way .

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

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

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

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

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

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

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