INTEGRALS RELATED TO THE LAPLACE TRANSFORM OF THE LOGARITHM

Let´s proof today the following integrals

$\int_0^{\infty}e^{-x}\ln x dx=-\gamma$

$\int_0^{\infty} e^{-x}\ln^2 x dx=\frac{\pi^2}{6}+\gamma^2$

$\int_0^{\infty} e^{-x}\ln^3 x dx=-2\zeta(3)-\frac{\gamma \pi^2}{2}-\gamma^3$

$\int_0^{\infty} e^{-x}\ln^4 x dx=\frac{3\pi^4}{20}+8\zeta(3)\gamma+\gamma^2\pi^2+\gamma^{4}$

Recall the integral representation of the Gamma Function

$\Gamma\left( z\right)=\int_0^{\infty}x^{z-1}e^{-x}dx$

We can differentiate under the integral sign with respect to to obtain the following results:

$\Gamma^{\prime}\left( z\right)=\int_0^{\infty} x^{z-1}e^{-x}\ln x dx$

$\Gamma^{\prime \prime}\left( z\right)=\int_0^{\infty} x^{z-1}e^{-x}\ln^2 x dx$

$\Gamma^{\prime \prime \prime}\left( z\right)=\int_0^{\infty} x^{z-1}e^{-x}\ln^3 x dx$

$\Gamma^{(4)}\left( z\right)=\int_0^{\infty} x^{z-1}e^{-x}\ln^4 x dx$

setting z=1 we obtain

$\Gamma^{\prime}\left( 1\right)=\int_0^{\infty}e^{-x}\ln x dx$ (1)

$\Gamma^{\prime \prime}\left( 1\right)=\int_0^{\infty} e^{-x}\ln^2 x dx$ (2)

$\Gamma^{\prime \prime \prime}\left( 1\right)=\int_0^{\infty} e^{-x}\ln^3 x dx$ (3)

$\Gamma^{(4)}\left( 1\right)=\int_0^{\infty} e^{-x}\ln^4 x dx$ (4)

On the other hand, we can obtain expressions for the derivatives of the Logarithm of the Gamma function as following

$\psi(x)=\frac{\Gamma^{\prime}(x)}{\Gamma(x)}$

$\boxed{\Gamma^{\prime}(x)=\Gamma(x)\psi(x)}$ (5)

$\Gamma^{\prime\prime}(x)=\Gamma^{\prime}(x)\psi(x)+\psi^{\prime}(x)\Gamma(x)$

$\Gamma^{\prime\prime}(x)=\frac{\Gamma(x)}{\Gamma(x)}\Gamma^{\prime}(x)\psi(x)+\psi^{\prime}(x)\Gamma(x)$

$\boxed{\Gamma^{\prime\prime}(x)=\Gamma(x)\psi^2(x)+\Gamma(x)\psi^{\prime}(x)}$ (6)

$\Gamma^{\prime\prime\prime}(x)=\Gamma^{\prime}(x)\psi^2(x)+2\Gamma(x)\psi(x)\psi^{\prime}(x)+\Gamma^{\prime}(x)\psi^{\prime}(x)+\Gamma(x)\psi^{\prime\prime}(x)$

$\Gamma^{\prime\prime\prime}(x)=\Gamma(x)\psi^3(x)+2\Gamma(x)\psi(x)\psi^{\prime}(x)+\Gamma(x)\psi(x)\psi^{\prime}(x)+\Gamma(x)\psi^{\prime\prime}(x)$

$\boxed{\Gamma^{\prime\prime\prime}(x)=\Gamma(x)\psi^3(x)+3\Gamma(x)\psi(x)\psi^{\prime}(x)+\Gamma(x)\psi^{\prime\prime}(x)}$ (7)

$\Gamma^{(4)}(x)=\Gamma(x)\psi^4(x)+3\Gamma(x)\psi^{2}(x)\psi^{\prime}(x)+3\left(\Gamma(x)\psi(x)^2\psi^{\prime}(x)+\Gamma(x)\left(\psi^{\prime}(x)\right)^2+\Gamma(x)\psi(x)\psi^{\prime\prime}(x)\right)+\Gamma(x)\psi(x)\psi^{\prime\prime}(x)+\Gamma(x)\psi^{\prime\prime \prime}(x)$

$\boxed{\Gamma^{(4)}(x)=\Gamma(x)\psi^4(x)+6\Gamma(x)\psi^{2}(x)\psi^{\prime}(x)+3\Gamma(x)\psi^{\prime}(x)^2+4\Gamma(x)\psi(x)\psi^{\prime\prime}(x)+\Gamma(x)\psi^{\prime\prime \prime}(x)}$ (8)

Now recall the special values of the polygamma function

$\psi\left( 1\right)=-\gamma$ (9)

$\psi^{\prime}\left( 1\right)=\zeta(2)$ (10)

$\psi^{\prime \prime}\left( 1\right)=-2\zeta(3)$ (11)

$\psi^{\prime \prime \prime}\left( 1\right)=6\zeta(4)$ (12)

Setting z=1 in (5) and using (9) we obtain

$\Gamma^{\prime}\left( 1\right)=\psi(1)\Gamma\left( 1\right)$

$\Gamma^{\prime}\left( 1\right)=-\gamma$ (13)

Equating (1) and (10) we get

$\boxed{\int_0^{\infty}e^{-x}\ln x dx=-\gamma}$ (14)

Setting z=1 in (6)

$\Gamma^{\prime \prime}\left( 1\right)=\left(\psi^{\prime}(1)+\psi^2(1)\right)\Gamma(1)$

$\Gamma^{\prime \prime}\left( 1\right)=\zeta(2)+\gamma^2$ (15)

Equating (2) and (15) we get

$\boxed{\int_0^{\infty} e^{-x}\ln^2 x dx=\frac{\pi^2}{6}+\gamma^2}$ (16)

Setting z=1 in (8)

$\Gamma^{\prime\prime\prime}(1)=\Gamma(1)\psi^3(1)+3\Gamma(1)\psi(1)\psi^{\prime}(1)+\Gamma(1)\psi^{\prime\prime}(1)$

$\Gamma^{\prime \prime \prime}\left( 1\right)=-2\zeta(3)-\frac{\gamma \pi^2}{2}-\gamma^3$ (17)

Equating (3) and (17) we get

$\boxed{\int_0^{\infty} e^{-x}\ln^3 x dx=-2\zeta(3)-\frac{\gamma \pi^2}{2}-\gamma^3}$ (18)

Setting z=1 in (7)

$\Gamma^{(4)}(1)=\Gamma(1)\psi^4(1)+6\Gamma(1)\psi^{2}(1)\psi^{\prime}(1)+3\Gamma(1)\psi^{\prime}(1)^2+4\Gamma(1)\psi(1)\psi^{\prime\prime}(1)+\Gamma(1)\psi^{\prime\prime \prime}(1)$

$\Gamma^{(4)}(1)=(-\gamma)^{4}+6(-\gamma)^2\zeta(2)+3\left(\zeta(2)\right)^2+4(-\gamma)\left(-2\zeta(3)\right)+6\zeta(4)$

$\Gamma^{(4)}(1)=\gamma^{4}+\gamma^2\pi^2+\frac{\pi^4}{12}+8\gamma\zeta(3)+\frac{\pi^2}{15}$

$\Gamma^{(4)}(1)=\frac{3\pi^4}{20}+8\zeta(3)\gamma+\gamma^2\pi^2+\gamma^{4}$ (19)

Equating (4) and (19) we get

$\boxed{\int_0^{\infty} e^{-x}\ln^4 x dx=\frac{3\pi^4}{20}+8\zeta(3)\gamma+\gamma^2\pi^2+\gamma^{4}}$ (20)

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INTEGRALS RELATED TO THE LAPLACE TRANSFORM OF THE LOGARITHM

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