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{\displaystyle \ln(a\cdot b)=\ln(a)+\ln(b)}
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{\displaystyle (4\pi \nu t)^{-1/2}}
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{\displaystyle u(x,t)=-2\nu {\frac {\partial }{\partial x}}\ln {\Bigl \{}(4\pi \nu t)^{-1/2}\int _{-\infty }^{\infty }\exp {\Bigl [}-{\frac {(x-x')^{2}}{4\nu t}}-{\frac {1}{2\nu }}\int _{0}^{x'}u(x'',0)dx''{\Bigr ]}dx'{\Bigr \}}.}
could be splitted from the integral. The differentiation
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{\displaystyle {\frac {\partial }{\partial x}}}
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{\displaystyle u(x,t)=-2\nu {\frac {\partial }{\partial x}}\ln {\Bigl \{}\int _{-\infty }^{\infty }\exp {\Bigl [}-{\frac {(x-x')^{2}}{4\nu t}}-{\frac {1}{2\nu }}\int _{0}^{x'}u(x'',0)dx''{\Bigr ]}dx'{\Bigr \}}}
which is simpler (but hides the origin of the solution, which is the heat equation). --Hero Wanders (talk ) 22:37, 14 March 2008 (UTC) [ reply ]
![{\displaystyle u(x,t)=-2\nu {\frac {\partial }{\partial x}}\ln {\Bigl \{}(4\pi \nu t)^{-1/2}\int _{-\infty }^{\infty }\exp {\Bigl [}-{\frac {(x-x')^{2}}{4\nu t}}-{\frac {1}{2\nu }}\int _{0}^{x'}u(x'',0)dx''{\Bigr ]}dx'{\Bigr \}}.}](https://wikimedia.org/enwiki/api/rest_v1/media/math/render/svg/320bc837fe94445e6aa36dc1db8a9f78c23f9854)
![{\displaystyle u(x,t)=-2\nu {\frac {\partial }{\partial x}}\ln {\Bigl \{}\int _{-\infty }^{\infty }\exp {\Bigl [}-{\frac {(x-x')^{2}}{4\nu t}}-{\frac {1}{2\nu }}\int _{0}^{x'}u(x'',0)dx''{\Bigr ]}dx'{\Bigr \}}}](https://wikimedia.org/enwiki/api/rest_v1/media/math/render/svg/46a1bb67871a439e9ec1e6939558f51eb507de83)