real value function (integral)

May 2010
39
0


How would I do this? Im really stuck. Most appreciated!
 
Mar 2010
280
122
\(\displaystyle \frac{df}{d \alpha} \: = \: \frac{1}{\alpha}\)
Integrating
\(\displaystyle f(\alpha) \: = \: ln | \alpha | +C\)
\(\displaystyle f(1) \: = \: C=0\)

\(\displaystyle f(\alpha \beta) \: = \: ln | \alpha | +ln | \beta |\)
\(\displaystyle f(\alpha \beta) \: = \: f(\alpha)+f(\beta) \)
 
May 2009
959
362
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

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

\(\displaystyle = \int^{\alpha}_{1} \frac{1}{t} \ dx = \ln \alpha \)
 
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simplependulum

MHF Hall of Honor
Jan 2009
715
427
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 \(\displaystyle f(ab) = f(a) + f(b)\)