gamma functionintegration
Here's something basic on the transformation of Gamma function and I'm wondering if someone could explain to me how the following works:
we know the function has two formulations:
$\Gamma(x) = \int_{0}^{\infty}t^{x-1}e^{-t}dt$
and
$\Gamma(x) = \int_{0}^{1}(-\log t)^{x-1}dt$
but how can we transform one into the other?
I've tried the method https://math.stackexchange.com/a/360687/573671 via user17762 but failed.
Thanks in advance for your time and help!
The real part of the integral is easily obtained thanks to the Fourier transform (see attachment). The imaginary part leads to much more difficulties. We even not know if a closed form exists.
The formula below shows the imaginary part expressed as a Cauchy Principal Value. I doubt that a simpler closed form could be derived.
The numerical tests are in very good agreement with this formula. Sorry, I don't presently publish the analytical calculus leading to the imaginary part because there is still a remaining theoretical difficulty.
[Typo corrected in the formula : 1/pi was missing ]
Your definition of incomplete Gamma function is wrong.
As an alternative definition of incomplete Gamma function: $$\Gamma(v,at)=\int^\infty_1(at)^vu^{v-1}e^{-atu}du=a^vt^v\int^\infty_1u^{v-1}e^{-atu}du$$
Your original integral $I$ becomes $$I=\int^\infty_0a^vt^v e^{-pt}\int^\infty_1u^{v-1}e^{-atu}du dt$$
By Fubini's theorem (the integrand is always positive), $$I=a^v\int^\infty_1 u^{v-1} \underbrace{\int^\infty_0 t^ve^{-pt}e^{-atu}dt}_{=I_1} du$$
$$I_1=\int^\infty_0t^ve^{-(p+au)t}dt$$
Let $g=(p+au)t$, $$I_1=\int^\infty_0\frac{g^v}{(p+au)^{v+1}}e^{-g}dg=\frac{\Gamma(v+1)}{(p+au)^{v+1}}$$
$$I=a^v\int^\infty_1 u^{v-1}\frac{\Gamma(v+1)}{(p+au)^{v+1}}du=a^v\Gamma(v+1)\int^\infty_1\left(\frac{u}{p+au}\right)^{v+1}\frac1{u^2}du$$
Substitute $h=\frac1u$, $$I=a^v\Gamma(v+1)\int^1_0\frac1{(ph+a)^{v+1}}dh=a^v\Gamma(v+1)\cdot\frac{-1}{pv}\left((p+a)^{-v}-a^{-v}\right)$$
By recognizing $\Gamma(v+1)=v\Gamma(v)$ and further simplifying, $$I=\frac{\Gamma(v)}p\left(1-\left(1+\frac pa\right)^{-v}\right)$$
An integral transform of Gamma function rarely exist, because Gamma function grows too rapidly. It grows even faster than exponential growth, that's why it does not have a Laplace transform. I'm not sure if a kernel like $e^{-x^2}$ would be okay.
Note that when we try to do a Laplace transform of Gamma function (incomplete or complete) with respect to the first argument, we always fail. If we do it w.r.t. to the second argument (for incomplete Gamma function), we get something useful, which is listed in your table.
Due to $$\lim_{x\to\infty}\frac{\Gamma(x)x^\alpha}{\Gamma(x+\alpha)}=1$$, I can 'invent' a kernel such that the transform for Gamma function exists.
$$\mathcal{T}_\alpha\{f\}(s)=\int^\infty_0 f(t)\cdot\frac{t^{\alpha-s}}{\Gamma(t+\alpha)}dt$$
$\mathcal{T}_\alpha\{\Gamma\}(s)$ exists if one of the following condition is satisfied:
Best Answer
They used the transformation $ t = -\log(u)$. Then you should note that $$ e^{-t}\,\mathrm{d}t = - \,\mathrm{d} u $$ Moreover, the integration bounds change from $(0,\infty) \mapsto (1,0)$. Now, you should reverse the order of integration to get the bounds $(0,1)$. This gives a minus sign which cancels against the one we had above.
I hope this helps. If you have any further questions, feel free to ask them!