More general Frullani's [closed]

$$\int_0^{\infty}\frac {(1+ax)^{-p} - (1+bx)^{-q}}{x}dx$$

• Attet: I tried to solve this problem by applying Frullani's theorem but failed suggestions are welcome

I don't understand why downvote when questions like Proof of Frullani's theorem are so appreciated.

Solution 1:

Let $f$ and $g$ be continuous functions on $[0,\infty)$. Assume that $$f(x) - f(\infty) = \sum_{k\ge0}\frac {u(k)(-x)^k}{k!} \;\text{ and }\; g(x) - g(\infty) = \sum_{k\ge0}\frac {v(k)(-x)^k}{k!},$$ where $$f(\infty) = \lim_{x\to\infty}f(x) \;\text{ and }\; g(\infty) = \lim_{x\to\infty}g(x)$$ exist and are finite.

The following statement is called Master Theorem (I don't remember exactly the source exactly but it was given by Hardy.): \begin{align*} I_n &= \int_0^{\infty}x^{n-1}(f(ax) - f(\infty)) - (g(bx) - g(\infty))dx \\ &=\Gamma(n)[a^{-n}(-n) - b^{-n}v(-n)]\\ &= \Gamma(n+1)\left(\frac {a^{-n}u(-n) - b^{-n}v(-n)}{n}\right), \end{align*} for $0<n<1$. We take the limit as $n\to 0^+$, hence \begin{align*} \lim_{n\to0^+}\left(\frac{b^nv(n) - a^u(n)}{n}\right) &= \lim_{n\to0^+}(n^nv(n)\ln b + b^nv'(n) - a^nu(n)\ln a - a^nu'(n)) \\ & = (f(0) - f(\infty))\left(\ln \frac ba + \frac d{ds}\left( \ln \left(\frac {v(s)}{u(s)}\right)_{s = 0}\right)\right). \end{align*} Here, $u(0) = v(0) = f(0) - f(\infty)$. Thus, $$I_0 = (f(0) - f(\infty))\left(\ln\left(\frac ba\right) + \frac d{ds}\left(\ln\left(\frac {v(s)}{u(s)}\right)\right)_{s = 0}\right).$$

In particular, $$\begin{align*} \int_0^{\infty}\frac {(1+ax)^{-p} - (1+bx)^{-q}}{x}dx & = \ln\frac ba + \frac {d}{ds} \ln\left(\frac {\Gamma(q+s)\Gamma(p)}{\Gamma(p+s)\Gamma (q)}\right)_{s =0}\\ & = \ln \frac ba + \psi(q) - \psi(p), \end{align*}$$ with $a, b, p, q > 0$.