# Thread: real value function (integral)

1. ## real value function (integral)

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

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

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

3. 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

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

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

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 .

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

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