# How to show $\int_{0}^{\infty}e^{-x}\ln^{2}x\:\mathrm{d}x=\gamma ^{2}+\frac{\pi ^{2}}{6}$?

• Elementary derivation of certian identites related to the Riemannian Zeta function and the Euler-Mascheroni Constant

#### Solutions Collecting From Web of "How to show $\int_{0}^{\infty}e^{-x}\ln^{2}x\:\mathrm{d}x=\gamma ^{2}+\frac{\pi ^{2}}{6}$?"

As already mentioned, the integral evaluates to $\Gamma”(1)$.

Recall the digamma and trigamma functions:

$$\psi_{0} (x) = \frac{d}{dx} \ln \Gamma(x) = \frac{\Gamma'(x)}{\Gamma(x)}$$

$$\psi_{1}(x) = \frac{d}{dx} \psi_{0}(x) = \frac{\Gamma”(x)\Gamma(x)- (\Gamma'(x))^{2}}{\Gamma^{2}(x)}$$

Then

$$\psi_{1}(1) = \frac{\Gamma”(1)\Gamma(1) + (\Gamma'(1))^{2}}{\Gamma^{2}(1)} = \Gamma”(1) – (\Gamma'(1))^{2}$$

The digamma and trigamma functions also have series definitions (which can be derived from the infinite product representation of the gamma function):

$$\psi_{0}(x) = – \gamma + \sum_{n=0}^{\infty} \Big( \frac{1}{n+1} – \frac{1}{n+x}\Big)$$

$$\psi_{1}(x) = \sum_{n=0}^{\infty} \frac{1}{(n+x)^{2}}$$

So

$$\psi_{1}(1) = \sum_{k=0}^{\infty} \frac{1}{(k+1)^2} = \zeta(2) = \frac{\pi^{2}}{6}$$

And

$$\Gamma'(1) = \psi_{0}(1) \Gamma(1) = \psi_{0}(1) = – \gamma + \sum_{n=0}^{\infty} \left(\frac{1}{n+1} – \frac{1}{n+1} \right) = – \gamma$$

Therefore,

$$\Gamma”(1) = \psi_{1}(1) + (\Gamma'(1))^{2} = \frac{\pi^{2}}{6} + \gamma^{2}$$

For Real-Valued $s>0$, consider the well-known $$\Gamma(s)=\int_{0}^{\infty}{t^{s-1}e^{-t}dt}$$ Now differentiating twice with respect to $s$ and by the Dominated convergence theorem, which allows us to differentiate inside the integral we have that $$\frac{\partial^{2}}{\partial s^{2}}\Gamma(s)=\int_{0}^{\infty}{ln^{2}(t)\cdot t^{s-1}e^{-t}dt}$$ plugging in $s=1$ yields our disired result.
The derivatives of the gamma function can be expressed as polygamma functions, where $$\frac{\Gamma'(s)}{\Gamma(s)}=\psi(s)$$

Related problems: I, II. Recalling the Mellin transform of a function $f$

$$F(s) = \int_{0}^{\infty}x^{s-1} f(x)dx$$

which gives

$$F”(s) = \int_{0}^{\infty}x^{s-1}\ln(x)^2 f(x) dx .$$

Now, for your case, taking $f(x)=e^{-x}$ which has its Mellin transform equals to $F(s)=\Gamma(s)$, where $\Gamma(s)$ is the gamma function, gives

$$F”(s) = \Gamma”(s) \longrightarrow (*) .$$

Taking the limit of $(*)$ as $s\to 1$ gives the desired result.

Note:

$$\Gamma'(s) = \Gamma(s)\psi(s).$$

Hint: Integration by parts.

$$f(x)=\ln^2 x$$
$$g^{\prime}(x)=e^{-x}$$