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I was answering this question: $\int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx$ and in my answer, I encountered the integral

$$\int_0^{\infty} \log^2(x) e^{-kx}dx$$ which according to WolframAlpha for $k=1,2,3$ and thereby generalizing gives us

$$\int_0^{\infty} \log^2(x) e^{-kx}dx = \dfrac{\pi^2}{6k} + \dfrac{(\gamma+ \ln(k))^2}{k} \tag{$\star$}$$

Also, in general, is there a nice form for

$$\int_0^{\infty} \log^m(x) e^{-kx} dx \tag{$\perp$}$$

EDIT

For $(\star)$, if we let $kx = t$, we then get $$I = \int_0^{\infty} \log^2(t/k) e^{-t} \dfrac{dt}k = \dfrac1k\int_0^{\infty}\log^2(t) e^{-t}dt-\dfrac{2\log(k)}{k}\int_0^{\infty}\log(t) e^{-t} dt + \dfrac{\log^2(k)}k$$

Considering the above, $(\star)$ and $(\perp)$ boil down to evaluating the integral

$$\color{red}{I(m) = \int_0^{\infty} \log^m(x) e^{-x} dx} \tag{$\spadesuit$}$$

Some values of $I(m)$. \begin{align} I(1) & = -\gamma\\ I(2) & = \zeta(2) + \gamma^2\\ I(3) & = -2\zeta(3) - 3 \gamma \zeta(2) - \gamma^3\\ I(4) & = \dfrac{27}2 \zeta(4) +8 \gamma \zeta(3) + 6 \gamma^2 \zeta(2) + \gamma^4\\ I(5) & = -24 \zeta(5) - \dfrac{135}2 \gamma \zeta(4) - 20 \gamma^2 \zeta(3) - 20\zeta(2) \zeta(3) - 10 \gamma^3 \zeta(2) -\gamma^5 \end{align}

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  • @GitGud I do not quite follow your second comment. Feel free to edit to change if something is unclear. –  May 17 '13 at 06:22
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    These numbers seem very closely related to the Laurent expansion of the Gamma function about $0$, but I don't know why this is the case or how to use this. – davidlowryduda May 17 '13 at 06:23
  • @GitGud Oh ok. Understood. Thanks. –  May 17 '13 at 06:28
  • @user17762 No problem. – Git Gud May 17 '13 at 06:28
  • For the second, your integral is simply $ \Gamma'(1) $ which is well known to be $ -\gamma $. The first one can be expressed using the double derivative of the Gamma function, which might help find resources on it. – Jon Claus May 17 '13 at 06:32
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    $$\int_0^{\infty} \log(x)^m e^{-x} dx = \left.\frac{\partial^m}{\partial\alpha^m} \int_0^{\infty} x^{\alpha} e^{-x} dx \right|_{\alpha=0} = \Gamma^{(m)}(1)$$ – achille hui May 17 '13 at 06:50
  • @achillehui Very nice. That qualifies to be answer! Could you also outline the method to obtain the $m^{th}$ derivatives of $\Gamma(1)$ in terms of $\gamma$ and the $\zeta$ function evaluated at the first $m$ values? –  May 17 '13 at 06:51
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    Start with Weierstrass infinite product definition of Gamma function:$$\frac{1}{\Gamma(z)} = e^{\gamma z}z\prod_{n=1}^{\infty}(1+\frac{z}{n})e^{-\frac{z}{n}}$$ It is not hard to derive following expansion of $\log \Gamma(z+1)$: $$\log\Gamma(z+1) = -\gamma z + \sum_{k=2}^{\infty} \frac{\zeta(k)}{k} (-z)^k$$ in terms of zeta functions. Taking exp and Taylor expand it allow you to express your integrals in terms of $\gamma$ and the zeta functions. – achille hui May 17 '13 at 07:13
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    A related problem. – Mhenni Benghorbal May 17 '13 at 07:42
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    A related problem. – Mhenni Benghorbal May 17 '13 at 07:51
  • @achillehui: that series for $\log(\Gamma(z+1))$ can also be easily derived from $(\ast)$ in my answer. – robjohn May 17 '13 at 12:59

2 Answers2

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As mentioned in this answer $$ \int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t=\Gamma'(1)=-\gamma $$ Using the definition $$ \Gamma(x)=\int_0^\infty t^{x-1}\,e^{-t}\,\mathrm{d}t $$ and taking the derivative $n$ times, we get $$ \Gamma^{(n)}(1)=\int_0^\infty\log(t)^n\,e^{-t}\,\mathrm{d}t $$ In the aforementioned answer, we also have $$ \Gamma'(x+1)=\Gamma(x+1)\left(-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\right)\tag{$\ast$} $$ Taking the derivative of $(\ast)$ at $x=0$ yields $$ \begin{align} \Gamma''(1) &=\Gamma'(1)\left(-\gamma\right)+\Gamma(1)\zeta(2)\\ &=\gamma^2+\zeta(2) \end{align} $$ Taking the second derivative of $(\ast)$ at $x=0$ yields $$ \begin{align} \Gamma'''(1) &=\Gamma''(1)(-\gamma)+2\Gamma'(1)\zeta(2)+\Gamma(1)(-2\zeta(3))\\ &=-\gamma^3-3\gamma\zeta(2)-2\zeta(3) \end{align} $$ We can use Leibniz rule and $$ \frac{\mathrm{d}^n}{\mathrm{d}x^n}\left(-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\right)=(-1)^{n+1}n!\,\zeta(n+1)\quad\text{ for }n\ge1 $$ applied to $(\ast)$, to get the recursion $$ \Gamma^{(n)}(1)=-\gamma\,\Gamma^{(n-1)}(1)+(n-1)!\sum_{k=2}^n(-1)^k\zeta(k)\frac{\Gamma^{(n-k)}(1)}{(n-k)!} $$

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robjohn
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  • I shall go through them, but I assume (?) in general $$\int_0^{\infty} \log^m(x) e^{-x} dx = \Gamma^{(m)}(1)$$ Could you also outline the method to obtain the $m^{th}$ derivatives of $\Gamma(1)$ in terms of $\gamma$ and the $\zeta$ function evaluated at the first $m$ values? –  May 17 '13 at 06:50
  • @user17762: yes. I should have added that to my answer (as I have now :-) – robjohn May 17 '13 at 06:57
  • Very nice. Thanks. –  May 17 '13 at 06:58
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Another method involving Mellin Transforms.

[This is my first time using this tactic so please correct me if I make a mistake, I am answering this question in part as practice]

Consider $F(s)$ as the Mellin Transform of $f(x)$

$$F(s)=\int_0^\infty x^{s-1} f(x)\text{ d}x$$

And thus

$$F''(s)=\int_0^\infty x^{s-1}\ln^2 x \space f(x) \text{ d}x$$

We assume that $f(x)=e^{-kx}$ and we are left with a nice known identity:

$$F(s)=\int_0^\infty x^{s-1}e^{-kx}=\frac{\Gamma(s)}{k^{s+1}}$$

Thus the answer to our integral is truly:

$$\lim_{s\to 3} \left(\left[\frac{\Gamma(s)}{k^{s-1}}\right]''\right)$$

Unfortunately, I am unable to calculate this limit, but to me, if you wanted to take a route similar to this, it seems much cleaner in the long run.

  • Please let me know if the limit evaluates to the result, but in theory it does. –  Mar 12 '16 at 06:20