Γ ( a , x ) = ∫ x ∞ t a − 1 exp ( − t ) d t Q ( a , x ) = 1 / Γ ( a ) ∫ x ∞ t a − 1 exp ( − t ) d t {\displaystyle {\begin{aligned}\Gamma (a,x)&=\int _{x}^{\infty }t^{a-1}\exp(-t)dt\\Q(a,x)&=1/\Gamma (a)\int _{x}^{\infty }t^{a-1}\exp(-t)dt\end{aligned}}}
Change (28) to Indefinite Integral
y = A ( t ) exp ∫ ∂ x x + B ( t ) exp [ 1 σ ¯ ∫ ∂ log x ] = A ( t ) exp ∫ x − 1 ∂ x 1 + B ( t ) x − 1 { exp [ ∫ ∂ log x ] } 1 σ ¯ = A ( t ) exp ∫ x − 1 ∂ x 1 + B ( t ) x − 1 { exp ( log x ) } 1 σ ¯ = A ( t ) exp ∫ x − 1 d x 1 + B ( t ) x 1 σ ¯ − 1 = A ( t ) exp ∫ d ( log x ) 1 + B ( t ) exp [ ( 1 σ ¯ − 1 ) log x ] {\displaystyle {\begin{aligned}y&=A(t)\exp \int {\frac {\partial x}{x+B(t)\exp[{\frac {1}{\bar {\sigma }}}\int \partial \log x]}}\\&=A(t)\exp \int {\frac {x^{-1}\partial x}{1+B(t)x^{-1}\{\exp[\int \partial \log x]\}^{\frac {1}{\bar {\sigma }}}}}\\&=A(t)\exp \int {\frac {x^{-1}\partial x}{1+B(t)x^{-1}\{\exp(\log x)\}^{\frac {1}{\bar {\sigma }}}}}\\&=A(t)\exp \int {\frac {x^{-1}dx}{1+B(t)x^{{\frac {1}{\bar {\sigma }}}-1}}}\\&=A(t)\exp \int {\frac {d(\log x)}{1+B(t)\exp[({\frac {1}{\bar {\sigma }}}-1)\log x]}}\end{aligned}}}
Let w = log x {\displaystyle w=\log x} , then the formular turns to be
y = A ( t ) exp ∫ d w 1 + B ( t ) exp [ ( 1 σ ¯ − 1 ) w ] {\displaystyle y=A(t)\exp \int {\frac {dw}{1+B(t)\exp[({\frac {1}{\bar {\sigma }}}-1)w]}}}
On the other hand, we have (adapted from the Integral Formula Table)
∫ d x a + b e m x = 1 a m [ m x − log ( a + b e m x ) ] {\displaystyle \int {\frac {dx}{a+be^{mx}}}={\tfrac {1}{am}}[mx-\log(a+be^{mx})]}
In this case, a = 1 {\displaystyle a=1} , b = B ( t ) {\displaystyle b=B(t)} and m = 1 σ ¯ − 1 {\displaystyle m={\frac {1}{\bar {\sigma }}}-1} , use the formula above, we get
y ( x , t ) = A ( t ) exp { 1 1 σ ¯ − 1 [ ( 1 σ ¯ − 1 ) w − log ( 1 + B ( t ) e 1 ( σ ¯ − 1 ) w ) ] } = A ( t ) exp { 1 1 σ ¯ − 1 [ ( 1 σ ¯ − 1 ) log x − log ( 1 + B ( t ) e 1 ( σ ¯ − 1 ) log x ) ] } = A ( t ) exp { − 1 1 σ ¯ − 1 [ − ( 1 σ ¯ − 1 ) log x + log ( 1 + B ( t ) x ( 1 σ ¯ − 1 ) ) ] } = A ( t ) [ exp { − log x 1 σ ¯ − 1 + log ( 1 + B ( t ) x 1 ( σ ¯ − 1 ) ) } ] − 1 1 σ ¯ − 1 = A ( t ) [ exp { log ( 1 + B ( t ) x 1 ( σ ¯ − 1 ) x 1 σ ¯ − 1 ) } ] − σ ¯ 1 − σ ¯ = A ( t ) ( x − ( 1 − σ ¯ σ ¯ ) + B ( t ) ) − ( σ ¯ 1 − σ ¯ ) {\displaystyle {\begin{aligned}y(x,t)&=A(t)\exp\{{\frac {1}{{\frac {1}{\bar {\sigma }}}-1}}[({\frac {1}{\bar {\sigma }}}-1)w-\log(1+B(t)e^{{\frac {1}{({\bar {\sigma }}}}-1)w})]\}\\&=A(t)\exp\{{\frac {1}{{\frac {1}{\bar {\sigma }}}-1}}[({\frac {1}{\bar {\sigma }}}-1)\log x-\log(1+B(t)e^{{\frac {1}{({\bar {\sigma }}}}-1)\log x})]\}\\&=A(t)\exp\{{\frac {-1}{{\frac {1}{\bar {\sigma }}}-1}}[-({\frac {1}{\bar {\sigma }}}-1)\log x+\log(1+B(t)x^{({\frac {1}{\bar {\sigma }}}-1)})]\}\\&=A(t)[\exp\{-\log x^{{\frac {1}{\bar {\sigma }}}-1}+\log(1+B(t)x^{{\frac {1}{({\bar {\sigma }}}}-1)})\}]^{\frac {-1}{{\frac {1}{\bar {\sigma }}}-1}}\\&=A(t)[\exp\{\log({\frac {1+B(t)x^{{\frac {1}{({\bar {\sigma }}}}-1)}}{x^{{\frac {1}{\bar {\sigma }}}-1}}})\}]^{\frac {-{\bar {\sigma }}}{1-{\bar {\sigma }}}}\\&=A(t)(x^{-({\frac {1-{\bar {\sigma }}}{\bar {\sigma }}})}+B(t))^{-({\frac {\bar {\sigma }}{1-{\bar {\sigma }}}})}\\\end{aligned}}}
![{\displaystyle {\begin{aligned}y&=A(t)\exp \int {\frac {\partial x}{x+B(t)\exp[{\frac {1}{\bar {\sigma }}}\int \partial \log x]}}\\&=A(t)\exp \int {\frac {x^{-1}\partial x}{1+B(t)x^{-1}\{\exp[\int \partial \log x]\}^{\frac {1}{\bar {\sigma }}}}}\\&=A(t)\exp \int {\frac {x^{-1}\partial x}{1+B(t)x^{-1}\{\exp(\log x)\}^{\frac {1}{\bar {\sigma }}}}}\\&=A(t)\exp \int {\frac {x^{-1}dx}{1+B(t)x^{{\frac {1}{\bar {\sigma }}}-1}}}\\&=A(t)\exp \int {\frac {d(\log x)}{1+B(t)\exp[({\frac {1}{\bar {\sigma }}}-1)\log x]}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1c54f0a45415a3cd7f1bb7f33b2a8bfa3d95a57)
, then the formular turns to be
![{\displaystyle y=A(t)\exp \int {\frac {dw}{1+B(t)\exp[({\frac {1}{\bar {\sigma }}}-1)w]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/646aeca3973b7a632b012fa4e4114b3fe7715aca)
![{\displaystyle \int {\frac {dx}{a+be^{mx}}}={\tfrac {1}{am}}[mx-\log(a+be^{mx})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d08bd6b9416acaff465240bbcd91298a10d64264)
,
and 
, use the formula above, we get
![{\displaystyle {\begin{aligned}y(x,t)&=A(t)\exp\{{\frac {1}{{\frac {1}{\bar {\sigma }}}-1}}[({\frac {1}{\bar {\sigma }}}-1)w-\log(1+B(t)e^{{\frac {1}{({\bar {\sigma }}}}-1)w})]\}\\&=A(t)\exp\{{\frac {1}{{\frac {1}{\bar {\sigma }}}-1}}[({\frac {1}{\bar {\sigma }}}-1)\log x-\log(1+B(t)e^{{\frac {1}{({\bar {\sigma }}}}-1)\log x})]\}\\&=A(t)\exp\{{\frac {-1}{{\frac {1}{\bar {\sigma }}}-1}}[-({\frac {1}{\bar {\sigma }}}-1)\log x+\log(1+B(t)x^{({\frac {1}{\bar {\sigma }}}-1)})]\}\\&=A(t)[\exp\{-\log x^{{\frac {1}{\bar {\sigma }}}-1}+\log(1+B(t)x^{{\frac {1}{({\bar {\sigma }}}}-1)})\}]^{\frac {-1}{{\frac {1}{\bar {\sigma }}}-1}}\\&=A(t)[\exp\{\log({\frac {1+B(t)x^{{\frac {1}{({\bar {\sigma }}}}-1)}}{x^{{\frac {1}{\bar {\sigma }}}-1}}})\}]^{\frac {-{\bar {\sigma }}}{1-{\bar {\sigma }}}}\\&=A(t)(x^{-({\frac {1-{\bar {\sigma }}}{\bar {\sigma }}})}+B(t))^{-({\frac {\bar {\sigma }}{1-{\bar {\sigma }}}})}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07a08e0fd1d887a781d715de7a06ef63d5d9503c)
