how to solve the complex logarithm problem with limit Re(z) converges to infinity?











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Prove that $$lim_{Re(z)rightarrow infty }(logGamma (z+1)-(z+frac{1}{2})log z+z)=frac{1}{2}log2pi.$$






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Prove that $$lim_{Re(z)rightarrow infty }(logGamma (z+1)-(z+frac{1}{2})log z+z)=frac{1}{2}log2pi.$$






analysis







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    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
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    Are you allowed the use of the Stirling formula ?
    – Yves Daoust
    20 hours ago













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Prove that $$lim_{Re(z)rightarrow infty }(logGamma (z+1)-(z+frac{1}{2})log z+z)=frac{1}{2}log2pi.$$






analysis







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Prove that $$lim_{Re(z)rightarrow infty }(logGamma (z+1)-(z+frac{1}{2})log z+z)=frac{1}{2}log2pi.$$






analysis




analysis






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edited 20 hours ago









Tianlalu

2,315632




2,315632






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  • 3




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    20 hours ago






  • 1




    Are you allowed the use of the Stirling formula ?
    – Yves Daoust
    20 hours ago














  • 3




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    20 hours ago






  • 1




    Are you allowed the use of the Stirling formula ?
    – Yves Daoust
    20 hours ago








3




3




Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
20 hours ago




Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
20 hours ago




1




1




Are you allowed the use of the Stirling formula ?
– Yves Daoust
20 hours ago




Are you allowed the use of the Stirling formula ?
– Yves Daoust
20 hours ago















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