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I have to show that for $\Re(\mu)>0$, $$\int_0^\infty x^2e^{-\mu x^2}\ln x\, \mathrm dx=\frac{1}{8\mu}(2-\ln 4\mu-C)\sqrt{\frac{\pi}{\mu}}$$

Source - Table of integrals, series, and products by I S Gradshteĭn on page number $605$

My attempt - \begin{align}\int_0^\infty x^2 e^{-\mu x^2}\ln x\, \mathrm dx&=\frac{1}{2}\int_0^\infty \sqrt{x}e^{-\mu x}\ln x\, \mathrm dx\, \text{ ,via substituting $x^2\mapsto x$}\end{align} Now, the Laplace Transform of $\ln x$ is $f(\mu)=-\frac{\ln(\mu)+\gamma}{\mu}$. And we have the result that $$\mathcal{L}(x^n \ln x)=(-1)^n f^n(\mu)$$ And here $n=\frac{1}{2}$. So, $$\mathcal{L}(x^{1/2}\ln x)=(-1)^{1/2}f^{1/2}(\mu)$$ How can I find the half derivative? It seems extremely absurd. Wolfram Alpha nth derivative calculator also says 'input not valid'.

Questions :

$1.$ Is my approach valid? Is my approach not yielding something?

$2.$ Are there any other approaches to solve this question?

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  • $\begingroup$ Hi, what is C? Can you help your post be more inclusive? $\endgroup$ Commented Jun 9, 2021 at 4:14
  • $\begingroup$ So why don't just integrate by parts? $\endgroup$ Commented Jun 9, 2021 at 4:21
  • $\begingroup$ It's going to be too lengthy. $\endgroup$
    – user730361
    Commented Jun 9, 2021 at 4:22
  • $\begingroup$ Yeah, if you say so. $\endgroup$ Commented Jun 9, 2021 at 4:24
  • $\begingroup$ I have an approach which doesn't require Laplace transform. So, if you want, I can provide that. $\endgroup$ Commented Jun 9, 2021 at 4:53

2 Answers 2

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The method chosen by the author of the problem is valid, makes use of transforms, and leads to having to use fractional calculus. This solution is more straight forward.

Also note that older books, like I S Gradshteĭn's tome, uses $C$ as the Euler-gamma constant while others use the more common $\gamma$.

Consider the integral $$ I(s, \alpha) = \int_{0}^{\infty} e^{- s \, t} \, t^{\alpha -1} \, dt = \frac{\Gamma(\alpha)}{s^\alpha}. $$ Differentiate the integral with respect to $\alpha$ to obtain \begin{align} \int_{0}^{\infty} e^{-s \, t} \, \partial_{\alpha} (t^{\alpha -1}) \, dt &= \partial_{\alpha} \left(\frac{\Gamma(\alpha)}{s^\alpha}\right) \\ \int_{0}^{\infty} e^{-s \, t} \, t^{\alpha -1} \, \ln(t) \, dt &= \frac{\Gamma(\alpha)}{s^{\alpha}} \, (\psi(\alpha) - \ln(s)). \end{align} Now let $ t = x^2$ to obtain $$ \int_{0}^{\infty} e^{-s \, x^2} \, x^{2 \alpha -1} \, \ln(x) \, dx = \frac{\Gamma(\alpha)}{4 \, s^{\alpha}} \, (\psi(\alpha) - \ln(s)). $$ When $\alpha = 3/2$ the integral reduces to $$ \int_{0}^{\infty} e^{-s \, x^2} \, x^{2} \, \ln(x) \, dx = \frac{\Gamma(3/2)}{4 \, s^{3/2}} \, (\psi(3/2) - \ln(s)). $$ Since $\psi(3/2) = 2 - \gamma - \ln(4)$, where $\gamma$ is the Euler-gamma constant, then $$ \int_{0}^{\infty} e^{-s \, x^2} \, x^{2} \, \ln(x) \, dx = \frac{1}{8 \, s} \, \sqrt{\frac{\pi}{s}} \, (2 - \gamma - \ln(4s)). $$

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I will provide an alternative approach using differentiation under integral sign.

We start with defining a function $ f :\mathbb{R} \mapsto \mathbb{R}$

$$f(a)=\int_0^\infty e^{-\mu x^2}x^a\, \mathrm dx $$

Notive that

$$\left. \dfrac{\mathrm d}{\mathrm da}f(a)\right|_{a=2}=I$$

Substituting $\mu x^2 \mapsto x $,

$$\begin{align} f(a) &= \frac{1}{2{\mu}^{(a+1)/2}}\int_0^{\infty} e^{-x}x^{(a+1)/2-1}\, \mathrm dx \\ f(a) &= \dfrac{ \Gamma\left(\frac{a+1}{2}\right)}{2\mu^{\frac{a+1}{2}}} \\ f^{\prime}(a) &= \dfrac{ \Gamma\left(\frac{a+1}{2}\right)}{4\mu^{ \frac{a+1}{2}}} \left[\psi\left(\frac{a+1}{2}\right)-\ln(\mu) \right] \\ I &=\dfrac{\Gamma(3/2)}{4 \mu^{3 /2}} (\psi(3/2)-\ln{\mu}) \end{align}$$

Using the recurrence relation of Gamma function $ \Gamma(a+1)=a\Gamma(a)$ and $\Gamma(1/2)= \sqrt{\pi}$ , we get $\Gamma(3/2)= \frac{\sqrt{\pi}}{2}$.

Using the series representation of digamma function

$$\psi(z)=-\gamma+\sum_{k=1}^{\infty}\dfrac{1}{k}-\ \dfrac{1}{k+z-1}$$

Here, $\gamma$ is the Euler-Mascheroni constant. And the series representation of $\ln{2}$.

$$\ln{2}=\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{k}$$

We get $\psi(3/2)= 2-\gamma-\ln{4}$.

Finally putting all the values into the equation for $I$ and simplifying gives the result

$$\boxed{\boxed{\int_0^{\infty} e^{-\mu x^2} x^2 \ln{x}\, \mathrm dx = \dfrac{1}{8\mu}(2-\gamma-\ln(4\mu))\sqrt{\frac{\pi}{\mu}} }}$$

So it turns out that the $C$ in the result in the book denotes the Euler-Mascheroni constant.

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