- #1
julian
Gold Member
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Euler's constant is defined as ##\gamma = \lim_{n \rightarrow \infty} \left( 1 + {1 \over 2} + {1 \over 3} + \cdots + {1 \over n} - \ln n \right)##. This can be easily represented as an integral:
##
\gamma = \int_0^\infty \left( {1 \over 1 - e^{-x}} - {1 \over x} \right) e^{-x} dx \qquad Eq.1 .
##
I want to show that another expression for ##\gamma## is:
##
\gamma = - \int_0^\infty e^{-x} \ln x \; dx \qquad Eq.2 .
##
How do you get from Eq.1 to Eq.2 (or the other way around)?
At this wesite
http://mathworld.wolfram.com/Euler-MascheroniConstant.html
it lists 4 integral expressions for ##\gamma## : (4)-(7). I've proven that (5) follows from (4), and that (7) follows from (6). If you can connect (4) or (5) to either (6) or (7), then that will be an answer to my question.
Thanks.
##
\gamma = \int_0^\infty \left( {1 \over 1 - e^{-x}} - {1 \over x} \right) e^{-x} dx \qquad Eq.1 .
##
I want to show that another expression for ##\gamma## is:
##
\gamma = - \int_0^\infty e^{-x} \ln x \; dx \qquad Eq.2 .
##
How do you get from Eq.1 to Eq.2 (or the other way around)?
At this wesite
http://mathworld.wolfram.com/Euler-MascheroniConstant.html
it lists 4 integral expressions for ##\gamma## : (4)-(7). I've proven that (5) follows from (4), and that (7) follows from (6). If you can connect (4) or (5) to either (6) or (7), then that will be an answer to my question.
Thanks.