Representation of Gamma function on $0 < operatorname{Re } z < 1$












1












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I've read that for $0 < operatorname{Re }z < 1$ the Gamma function has the following representation:



$$Gamma(z) = e^{ipi frac{z}{2}} int_0^infty t^{z-1} e^{-it} , dt.$$
I couldn't find any proof of this, does someone have a reference? Your help is appreciated.






integration complex-analysis analysis gamma-function







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  • $begingroup$
    Let $x=it$$phantom{}$
    $endgroup$
    – Frank W.
    Jan 15 at 19:06










  • $begingroup$
    How is that step justified?
    $endgroup$
    – bavor42
    Jan 15 at 19:36
















1












$begingroup$


I've read that for $0 < operatorname{Re }z < 1$ the Gamma function has the following representation:



$$Gamma(z) = e^{ipi frac{z}{2}} int_0^infty t^{z-1} e^{-it} , dt.$$
I couldn't find any proof of this, does someone have a reference? Your help is appreciated.






integration complex-analysis analysis gamma-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    Let $x=it$$phantom{}$
    $endgroup$
    – Frank W.
    Jan 15 at 19:06










  • $begingroup$
    How is that step justified?
    $endgroup$
    – bavor42
    Jan 15 at 19:36














1












1








1





$begingroup$


I've read that for $0 < operatorname{Re }z < 1$ the Gamma function has the following representation:



$$Gamma(z) = e^{ipi frac{z}{2}} int_0^infty t^{z-1} e^{-it} , dt.$$
I couldn't find any proof of this, does someone have a reference? Your help is appreciated.






integration complex-analysis analysis gamma-function







share|cite|improve this question









$endgroup$




I've read that for $0 < operatorname{Re }z < 1$ the Gamma function has the following representation:



$$Gamma(z) = e^{ipi frac{z}{2}} int_0^infty t^{z-1} e^{-it} , dt.$$
I couldn't find any proof of this, does someone have a reference? Your help is appreciated.






integration complex-analysis analysis gamma-function




integration complex-analysis analysis gamma-function






share|cite|improve this question













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share|cite|improve this question










asked Jan 15 at 18:44









bavor42bavor42

315110




315110












  • $begingroup$
    Let $x=it$$phantom{}$
    $endgroup$
    – Frank W.
    Jan 15 at 19:06










  • $begingroup$
    How is that step justified?
    $endgroup$
    – bavor42
    Jan 15 at 19:36


















  • $begingroup$
    Let $x=it$$phantom{}$
    $endgroup$
    – Frank W.
    Jan 15 at 19:06










  • $begingroup$
    How is that step justified?
    $endgroup$
    – bavor42
    Jan 15 at 19:36
















$begingroup$
Let $x=it$$phantom{}$
$endgroup$
– Frank W.
Jan 15 at 19:06




$begingroup$
Let $x=it$$phantom{}$
$endgroup$
– Frank W.
Jan 15 at 19:06












$begingroup$
How is that step justified?
$endgroup$
– bavor42
Jan 15 at 19:36




$begingroup$
How is that step justified?
$endgroup$
– bavor42
Jan 15 at 19:36










1 Answer
1






active

oldest

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4












$begingroup$

For $R > r>0$ consider the contour consisting of the four paths
begin{align}
&gamma_1 colon [r,R] to mathbb{C} , , , t mapsto t , , \
&gamma_2 colon left[0,frac{pi}{2}right] to mathbb{C} , , , phi mapsto R mathrm{e}^{mathrm{i} phi} , , \
&gamma_3 colon [r,R] to mathbb{C} , , , s mapsto mathrm{i} (R + r - s) , , \
&gamma_4 colon left[0,frac{pi}{2}right] to mathbb{C} , , , phi mapsto r mathrm{e}^{mathrm{i} left(frac{pi}{2} - phi right)} , . \
end{align}
For $z in mathbb{C}$ with $operatorname{Re} (z) in (0,1)$ the function
$$ f colon mathbb{C}setminus mathbb{R}_{leq 0} to mathbb{C} , , , eta mapsto eta^{z-1} mathrm{e}^{- eta} , ,$$
is holomorphic, so
$$ sum limits_{k=1}^4 int limits_{gamma_k} f(eta) , mathrm{d} eta = 0 $$
holds by Cauchy's integral theorem. We have
begin{align}
left| , int limits_{gamma_2} f(eta) , mathrm{d} eta ,right| &= left| ,int limits_0^{pi/2} (R mathrm{e}^{mathrm{i} phi})^z mathrm{e}^{-R mathrm{e}^{mathrm{i} phi}} mathrm{i} , mathrm{d} phi , right| leq R^{operatorname{Re}(z)} mathrm{e}^{pi | operatorname{Im}(z)| /2} int limits_0^{pi/2} mathrm{e}^{- R cos(phi)} , mathrm{d} phi \
&leq R^{operatorname{Re}(z)} mathrm{e}^{pi | operatorname{Im}(z)| /2} int limits_0^{pi/2} mathrm{e}^{- R left(1-frac{2}{pi} phiright)} , mathrm{d} phi = frac{pi}{2} mathrm{e}^{pi | operatorname{Im}(z)| /2} (1 - mathrm{e}^{-R}) R^{-(1-operatorname{Re}(z))} stackrel{Rto infty}{longrightarrow} 0
end{align}
and, similarly,
$$ left| , int limits_{gamma_4} f(eta) , mathrm{d} eta ,right| leq frac{pi}{2} mathrm{e}^{pi | operatorname{Im}(z)| /2} (1 - mathrm{e}^{-r}) r^{-(1-operatorname{Re}(z))} stackrel{rto 0}{longrightarrow} 0 , .$$
Therefore
begin{align}
Gamma(z) &= int limits_0^infty f(t) , mathrm{d} t = lim_{R to infty} lim_{r to 0} int limits_{gamma_1} f(eta) , mathrm{d} eta = - lim_{R to infty} lim_{r to 0} int limits_{gamma_3} f(eta) , mathrm{d} eta \
&= - lim_{R to infty} lim_{r to 0} int limits_r^R [mathrm{i} (R+r-s)]^{z-1} mathrm{e}^{- mathrm{i} (R+r-s)} (- mathrm{i}) , mathrm{d} s \
&= i^{z} lim_{R to infty} lim_{r to 0} int limits_r^R t^{z-1} mathrm{e}^{-mathrm{i} t} , mathrm{d} t = mathrm{e}^{mathrm{i} frac{pi}{2} z} int limits_0^infty t^{z-1} mathrm{e}^{-mathrm{i} t} , mathrm{d} t , .
end{align}






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  • $begingroup$
    Really helpful answer. Thank you!
    $endgroup$
    – bavor42
    Jan 15 at 21:14











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1 Answer
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4












$begingroup$

For $R > r>0$ consider the contour consisting of the four paths
begin{align}
&gamma_1 colon [r,R] to mathbb{C} , , , t mapsto t , , \
&gamma_2 colon left[0,frac{pi}{2}right] to mathbb{C} , , , phi mapsto R mathrm{e}^{mathrm{i} phi} , , \
&gamma_3 colon [r,R] to mathbb{C} , , , s mapsto mathrm{i} (R + r - s) , , \
&gamma_4 colon left[0,frac{pi}{2}right] to mathbb{C} , , , phi mapsto r mathrm{e}^{mathrm{i} left(frac{pi}{2} - phi right)} , . \
end{align}
For $z in mathbb{C}$ with $operatorname{Re} (z) in (0,1)$ the function
$$ f colon mathbb{C}setminus mathbb{R}_{leq 0} to mathbb{C} , , , eta mapsto eta^{z-1} mathrm{e}^{- eta} , ,$$
is holomorphic, so
$$ sum limits_{k=1}^4 int limits_{gamma_k} f(eta) , mathrm{d} eta = 0 $$
holds by Cauchy's integral theorem. We have
begin{align}
left| , int limits_{gamma_2} f(eta) , mathrm{d} eta ,right| &= left| ,int limits_0^{pi/2} (R mathrm{e}^{mathrm{i} phi})^z mathrm{e}^{-R mathrm{e}^{mathrm{i} phi}} mathrm{i} , mathrm{d} phi , right| leq R^{operatorname{Re}(z)} mathrm{e}^{pi | operatorname{Im}(z)| /2} int limits_0^{pi/2} mathrm{e}^{- R cos(phi)} , mathrm{d} phi \
&leq R^{operatorname{Re}(z)} mathrm{e}^{pi | operatorname{Im}(z)| /2} int limits_0^{pi/2} mathrm{e}^{- R left(1-frac{2}{pi} phiright)} , mathrm{d} phi = frac{pi}{2} mathrm{e}^{pi | operatorname{Im}(z)| /2} (1 - mathrm{e}^{-R}) R^{-(1-operatorname{Re}(z))} stackrel{Rto infty}{longrightarrow} 0
end{align}
and, similarly,
$$ left| , int limits_{gamma_4} f(eta) , mathrm{d} eta ,right| leq frac{pi}{2} mathrm{e}^{pi | operatorname{Im}(z)| /2} (1 - mathrm{e}^{-r}) r^{-(1-operatorname{Re}(z))} stackrel{rto 0}{longrightarrow} 0 , .$$
Therefore
begin{align}
Gamma(z) &= int limits_0^infty f(t) , mathrm{d} t = lim_{R to infty} lim_{r to 0} int limits_{gamma_1} f(eta) , mathrm{d} eta = - lim_{R to infty} lim_{r to 0} int limits_{gamma_3} f(eta) , mathrm{d} eta \
&= - lim_{R to infty} lim_{r to 0} int limits_r^R [mathrm{i} (R+r-s)]^{z-1} mathrm{e}^{- mathrm{i} (R+r-s)} (- mathrm{i}) , mathrm{d} s \
&= i^{z} lim_{R to infty} lim_{r to 0} int limits_r^R t^{z-1} mathrm{e}^{-mathrm{i} t} , mathrm{d} t = mathrm{e}^{mathrm{i} frac{pi}{2} z} int limits_0^infty t^{z-1} mathrm{e}^{-mathrm{i} t} , mathrm{d} t , .
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Really helpful answer. Thank you!
    $endgroup$
    – bavor42
    Jan 15 at 21:14
















4












$begingroup$

For $R > r>0$ consider the contour consisting of the four paths
begin{align}
&gamma_1 colon [r,R] to mathbb{C} , , , t mapsto t , , \
&gamma_2 colon left[0,frac{pi}{2}right] to mathbb{C} , , , phi mapsto R mathrm{e}^{mathrm{i} phi} , , \
&gamma_3 colon [r,R] to mathbb{C} , , , s mapsto mathrm{i} (R + r - s) , , \
&gamma_4 colon left[0,frac{pi}{2}right] to mathbb{C} , , , phi mapsto r mathrm{e}^{mathrm{i} left(frac{pi}{2} - phi right)} , . \
end{align}
For $z in mathbb{C}$ with $operatorname{Re} (z) in (0,1)$ the function
$$ f colon mathbb{C}setminus mathbb{R}_{leq 0} to mathbb{C} , , , eta mapsto eta^{z-1} mathrm{e}^{- eta} , ,$$
is holomorphic, so
$$ sum limits_{k=1}^4 int limits_{gamma_k} f(eta) , mathrm{d} eta = 0 $$
holds by Cauchy's integral theorem. We have
begin{align}
left| , int limits_{gamma_2} f(eta) , mathrm{d} eta ,right| &= left| ,int limits_0^{pi/2} (R mathrm{e}^{mathrm{i} phi})^z mathrm{e}^{-R mathrm{e}^{mathrm{i} phi}} mathrm{i} , mathrm{d} phi , right| leq R^{operatorname{Re}(z)} mathrm{e}^{pi | operatorname{Im}(z)| /2} int limits_0^{pi/2} mathrm{e}^{- R cos(phi)} , mathrm{d} phi \
&leq R^{operatorname{Re}(z)} mathrm{e}^{pi | operatorname{Im}(z)| /2} int limits_0^{pi/2} mathrm{e}^{- R left(1-frac{2}{pi} phiright)} , mathrm{d} phi = frac{pi}{2} mathrm{e}^{pi | operatorname{Im}(z)| /2} (1 - mathrm{e}^{-R}) R^{-(1-operatorname{Re}(z))} stackrel{Rto infty}{longrightarrow} 0
end{align}
and, similarly,
$$ left| , int limits_{gamma_4} f(eta) , mathrm{d} eta ,right| leq frac{pi}{2} mathrm{e}^{pi | operatorname{Im}(z)| /2} (1 - mathrm{e}^{-r}) r^{-(1-operatorname{Re}(z))} stackrel{rto 0}{longrightarrow} 0 , .$$
Therefore
begin{align}
Gamma(z) &= int limits_0^infty f(t) , mathrm{d} t = lim_{R to infty} lim_{r to 0} int limits_{gamma_1} f(eta) , mathrm{d} eta = - lim_{R to infty} lim_{r to 0} int limits_{gamma_3} f(eta) , mathrm{d} eta \
&= - lim_{R to infty} lim_{r to 0} int limits_r^R [mathrm{i} (R+r-s)]^{z-1} mathrm{e}^{- mathrm{i} (R+r-s)} (- mathrm{i}) , mathrm{d} s \
&= i^{z} lim_{R to infty} lim_{r to 0} int limits_r^R t^{z-1} mathrm{e}^{-mathrm{i} t} , mathrm{d} t = mathrm{e}^{mathrm{i} frac{pi}{2} z} int limits_0^infty t^{z-1} mathrm{e}^{-mathrm{i} t} , mathrm{d} t , .
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Really helpful answer. Thank you!
    $endgroup$
    – bavor42
    Jan 15 at 21:14














4












4








4





$begingroup$

For $R > r>0$ consider the contour consisting of the four paths
begin{align}
&gamma_1 colon [r,R] to mathbb{C} , , , t mapsto t , , \
&gamma_2 colon left[0,frac{pi}{2}right] to mathbb{C} , , , phi mapsto R mathrm{e}^{mathrm{i} phi} , , \
&gamma_3 colon [r,R] to mathbb{C} , , , s mapsto mathrm{i} (R + r - s) , , \
&gamma_4 colon left[0,frac{pi}{2}right] to mathbb{C} , , , phi mapsto r mathrm{e}^{mathrm{i} left(frac{pi}{2} - phi right)} , . \
end{align}
For $z in mathbb{C}$ with $operatorname{Re} (z) in (0,1)$ the function
$$ f colon mathbb{C}setminus mathbb{R}_{leq 0} to mathbb{C} , , , eta mapsto eta^{z-1} mathrm{e}^{- eta} , ,$$
is holomorphic, so
$$ sum limits_{k=1}^4 int limits_{gamma_k} f(eta) , mathrm{d} eta = 0 $$
holds by Cauchy's integral theorem. We have
begin{align}
left| , int limits_{gamma_2} f(eta) , mathrm{d} eta ,right| &= left| ,int limits_0^{pi/2} (R mathrm{e}^{mathrm{i} phi})^z mathrm{e}^{-R mathrm{e}^{mathrm{i} phi}} mathrm{i} , mathrm{d} phi , right| leq R^{operatorname{Re}(z)} mathrm{e}^{pi | operatorname{Im}(z)| /2} int limits_0^{pi/2} mathrm{e}^{- R cos(phi)} , mathrm{d} phi \
&leq R^{operatorname{Re}(z)} mathrm{e}^{pi | operatorname{Im}(z)| /2} int limits_0^{pi/2} mathrm{e}^{- R left(1-frac{2}{pi} phiright)} , mathrm{d} phi = frac{pi}{2} mathrm{e}^{pi | operatorname{Im}(z)| /2} (1 - mathrm{e}^{-R}) R^{-(1-operatorname{Re}(z))} stackrel{Rto infty}{longrightarrow} 0
end{align}
and, similarly,
$$ left| , int limits_{gamma_4} f(eta) , mathrm{d} eta ,right| leq frac{pi}{2} mathrm{e}^{pi | operatorname{Im}(z)| /2} (1 - mathrm{e}^{-r}) r^{-(1-operatorname{Re}(z))} stackrel{rto 0}{longrightarrow} 0 , .$$
Therefore
begin{align}
Gamma(z) &= int limits_0^infty f(t) , mathrm{d} t = lim_{R to infty} lim_{r to 0} int limits_{gamma_1} f(eta) , mathrm{d} eta = - lim_{R to infty} lim_{r to 0} int limits_{gamma_3} f(eta) , mathrm{d} eta \
&= - lim_{R to infty} lim_{r to 0} int limits_r^R [mathrm{i} (R+r-s)]^{z-1} mathrm{e}^{- mathrm{i} (R+r-s)} (- mathrm{i}) , mathrm{d} s \
&= i^{z} lim_{R to infty} lim_{r to 0} int limits_r^R t^{z-1} mathrm{e}^{-mathrm{i} t} , mathrm{d} t = mathrm{e}^{mathrm{i} frac{pi}{2} z} int limits_0^infty t^{z-1} mathrm{e}^{-mathrm{i} t} , mathrm{d} t , .
end{align}






share|cite|improve this answer









$endgroup$



For $R > r>0$ consider the contour consisting of the four paths
begin{align}
&gamma_1 colon [r,R] to mathbb{C} , , , t mapsto t , , \
&gamma_2 colon left[0,frac{pi}{2}right] to mathbb{C} , , , phi mapsto R mathrm{e}^{mathrm{i} phi} , , \
&gamma_3 colon [r,R] to mathbb{C} , , , s mapsto mathrm{i} (R + r - s) , , \
&gamma_4 colon left[0,frac{pi}{2}right] to mathbb{C} , , , phi mapsto r mathrm{e}^{mathrm{i} left(frac{pi}{2} - phi right)} , . \
end{align}
For $z in mathbb{C}$ with $operatorname{Re} (z) in (0,1)$ the function
$$ f colon mathbb{C}setminus mathbb{R}_{leq 0} to mathbb{C} , , , eta mapsto eta^{z-1} mathrm{e}^{- eta} , ,$$
is holomorphic, so
$$ sum limits_{k=1}^4 int limits_{gamma_k} f(eta) , mathrm{d} eta = 0 $$
holds by Cauchy's integral theorem. We have
begin{align}
left| , int limits_{gamma_2} f(eta) , mathrm{d} eta ,right| &= left| ,int limits_0^{pi/2} (R mathrm{e}^{mathrm{i} phi})^z mathrm{e}^{-R mathrm{e}^{mathrm{i} phi}} mathrm{i} , mathrm{d} phi , right| leq R^{operatorname{Re}(z)} mathrm{e}^{pi | operatorname{Im}(z)| /2} int limits_0^{pi/2} mathrm{e}^{- R cos(phi)} , mathrm{d} phi \
&leq R^{operatorname{Re}(z)} mathrm{e}^{pi | operatorname{Im}(z)| /2} int limits_0^{pi/2} mathrm{e}^{- R left(1-frac{2}{pi} phiright)} , mathrm{d} phi = frac{pi}{2} mathrm{e}^{pi | operatorname{Im}(z)| /2} (1 - mathrm{e}^{-R}) R^{-(1-operatorname{Re}(z))} stackrel{Rto infty}{longrightarrow} 0
end{align}
and, similarly,
$$ left| , int limits_{gamma_4} f(eta) , mathrm{d} eta ,right| leq frac{pi}{2} mathrm{e}^{pi | operatorname{Im}(z)| /2} (1 - mathrm{e}^{-r}) r^{-(1-operatorname{Re}(z))} stackrel{rto 0}{longrightarrow} 0 , .$$
Therefore
begin{align}
Gamma(z) &= int limits_0^infty f(t) , mathrm{d} t = lim_{R to infty} lim_{r to 0} int limits_{gamma_1} f(eta) , mathrm{d} eta = - lim_{R to infty} lim_{r to 0} int limits_{gamma_3} f(eta) , mathrm{d} eta \
&= - lim_{R to infty} lim_{r to 0} int limits_r^R [mathrm{i} (R+r-s)]^{z-1} mathrm{e}^{- mathrm{i} (R+r-s)} (- mathrm{i}) , mathrm{d} s \
&= i^{z} lim_{R to infty} lim_{r to 0} int limits_r^R t^{z-1} mathrm{e}^{-mathrm{i} t} , mathrm{d} t = mathrm{e}^{mathrm{i} frac{pi}{2} z} int limits_0^infty t^{z-1} mathrm{e}^{-mathrm{i} t} , mathrm{d} t , .
end{align}







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share|cite|improve this answer



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answered Jan 15 at 20:55









ComplexYetTrivialComplexYetTrivial

4,6032631




4,6032631












  • $begingroup$
    Really helpful answer. Thank you!
    $endgroup$
    – bavor42
    Jan 15 at 21:14


















  • $begingroup$
    Really helpful answer. Thank you!
    $endgroup$
    – bavor42
    Jan 15 at 21:14
















$begingroup$
Really helpful answer. Thank you!
$endgroup$
– bavor42
Jan 15 at 21:14




$begingroup$
Really helpful answer. Thank you!
$endgroup$
– bavor42
Jan 15 at 21:14


















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in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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