Whilst messing around with the integral $$\int_a^b \Gamma(z)dz$$ I accidentally found the identity $$\int_a^b \Gamma(z)dz=\int_0^1 \frac{1}{z}\big[\Gamma(b+\ln(z))-\Gamma(a+\ln(z))\big]dz$$ But I am afraid that this identity may not be true, as when I tried to test it with $a=1$, $b=2$, Wolfram Alpha would not give me a value for the second part. Here is my derivation for this identity:
$$\int_a^b \Gamma(z)dz$$ $$=\int_a^b \int_0^\infty x^{z-1}e^{-x} dx dz$$ $$=\int_0^\infty \frac{x^{b-1}e^{-x}-x^{a-1}e^{-x}}{\ln(x)} dx dz$$
Now define the function $I(n)$ as $$I(n)=\int_0^\infty \frac{x^{b-1}e^{-x}-x^{a-1}e^{-x}}{\ln(x)}n^{\ln(x)} dx$$ So that $I(0)=0$ and $I(1)=\int_0^\infty \frac{x^{b-1}e^{-x}-x^{a-1}e^{-x}}{\ln(x)} dx dz$. Then we have $$I'(n)=\int_0^\infty (x^{b-1}e^{-x}-x^{a-1}e^{-x})n^{\ln(x)-1} dx$$ $$I'(n)=\frac{1}{n}\int_0^\infty (x^{b-1}e^{-x}-x^{a-1}e^{-x})n^{\ln(x)}dx$$ $$I'(n)=\frac{1}{n}\int_0^\infty (x^{b-1}e^{-x}-x^{a-1}e^{-x})x^{\ln(n)} dx$$ $$I'(n)=\frac{1}{n}\int_0^\infty x^{\ln(n)+b-1}e^{-x}-x^{\ln(n)+a-1}e^{-x} dx$$ $$I'(n)=\frac{1}{n}\int_0^\infty x^{\ln(n)+b-1}e^{-x}-x^{\ln(n)+a-1}e^{-x} dx$$ $$I'(n)=\frac{1}{n}\big[\Gamma(\ln(n)+b)-\Gamma(\ln(n)+a)\big]$$ And because of the previously stated values of $I(0)$ and $I(1)$, $$\int_a^b \Gamma(z)dz=I(1)-I(0)$$ $$\int_a^b \Gamma(z)dz=\int_0^1 I'(n) dn$$ $$\int_a^b \Gamma(z)dz=\int_0^1 \frac{1}{n}\big[\Gamma(\ln(n)+b)-\Gamma(\ln(n)+a)\big] dn$$
Can anyone expose the error in this derivation, or confirm that it is correct?
Very early on, I spot an error:
$$I(0)\ne0$$
You probably were thinking that $0^{\text{anything}}=0$, but if $\text{anything}<0$, then we have a problem.
On the other hand, the derivative doesn't look so bad:
$$I'(t)=\frac{\Gamma(b+\ln(t))-\Gamma(a+\ln(t))}t$$
Taking the anti-derivative is easy by splitting parts and letting $u=\ln(t)$.
$$\int I'(t)~\mathrm dt=\int\Gamma(b+u)-\Gamma(a+u)~\mathrm du$$
So you can probably see that it relates directly back to the original integral very simply.