Logarithmic Integrals?

Relevant wiki: Differentiation Under the Integral Sign

Let $a, b \in \mathbb{R}$ with $a,b > 1$ , and let $f(a,b) = \displaystyle\int_{1}^{\infty} \left(\dfrac{1}{x^a} - \dfrac{1}{x^b} \right) \dfrac{dx}{\ln(x)}$ . Then,

$\begin{array}{ccl} \dfrac{d(f(a,b))}{db} & = & \dfrac{d}{db}\left(\displaystyle\int_{1}^{\infty} \left(\dfrac{1}{x^a} - \dfrac{1}{x^b} \right)\dfrac{dx}{\ln(x)}\right) \\ & = & \displaystyle\int_{1}^{\infty} \dfrac{d}{db}\left(\dfrac{1}{x^a} - \dfrac{1}{x^b} \right)\dfrac{dx}{\ln(x)} \\ & = & \displaystyle\int_{1}^{\infty} \dfrac{1}{x^b} dx \\ & = & \dfrac{1}{b-1} \end{array}$

Therefore, $f(a,b) = \ln |b-1| + C$

Since $f(a,a) = 0$ , then $C = -\ln |a-1|$ . Hence, $f(a,b) = \displaystyle\int_{1}^{\infty} \left(\dfrac{1}{x^a} - \dfrac{1}{x^b} \right) \dfrac{dx}{\ln(x)} = \ln\left|\dfrac{b-1}{a-1}\right|$

Therefore, $e^{f(2,4)} = e^{\ln(3)} = \boxed{3}$ .

This is an instance of Frullani integral in disguise. Substitute $u = e^x$ . Then $A = \int_{0}^{\infty} \frac{e^{-x} - e^{-3x}}{x} \, \mathrm{d}x,$ and so, the answer is $\log 3$ . For those who are not familiar with Frullani integral, the trick is to write $\frac{e^{-x} - e^{-3x}}{x} = \int_{1}^{3} e^{-sx} \, \mathrm{d}s$ and interchange the order of integration, which is possibly by Fubini/Tonelli theorem: $A = \int_{0}^{\infty} \left( \int_{1}^{3} e^{-sx} \, \mathrm{d}s \right) \, \mathrm{d}x = \int_{1}^{3} \left( \int_{0}^{\infty} e^{-sx} \, \mathrm{d}x \right) \, \mathrm{d}s = \int_{1}^{3} \frac{\mathrm{d}s}{s} = \log 3.$

Put ln u = t, and use the Frullani Integral.

Another pedestrian approach with Differentiation Under the Integral Sign....

Let $u = e^{-x}$

$A = \int_{0}^{-\infty\!} (\frac{1}{e^{-2x}} - \frac{1}{e^{-4x}} ) \frac{-e^{-x}}{-x} dx$

Rearranging....

$A = \int_{-\infty\!}^{0} \frac{e^{3x}}{x}-\frac{e^x}{x}dx$

Let $I(n) = \int_{-\infty\!}^{0}\frac{e^{nx}}{x}dx$ so that $A = I(3) - I(1)$

Taking the derivative of $I(n)$ with respect to n yields

$I'(n) = \int_{-\infty\!}^{0}\frac{\partial}{\partial n}\frac{e^{nx}}{x}dx = \int_{-\infty\!}^{0}e^{nx}dx = \frac{1}{n}$

And re-integrating leads to

$I(n) = \int I'(n) dn = \int \frac{1}{n} dn = ln(n) + C$

And thus

$A = I(3) - I(1) = ln(3) + C - ln(1) - C = ln(3)$

Therefore $e^A = e^{ln(3)} = \fbox{3}$

Another possibility to apply Differentiation Under the Integral Sign :

$\begin{aligned} I(\alpha)&:=\int_1^\infty u^\alpha(u^{-a}-u^{-b})\frac{du}{\ln(u)}&&a,\:b>1,&&\alpha < \min\{a;\:b\}-1 \end{aligned}$

We want to calculate $I(0)$ and start with the derivative of $I(\alpha)$ :

$\begin{aligned} I^{(1)}(\alpha)&=\int_1^\infty u^{\alpha}(u^{-a}-u^{-b})\:du = \left[ \frac{u^{\alpha-a+1}}{\alpha-a+1} - \frac{u^{\alpha-b+1}}{\alpha-b+1} \right]_1^\infty\\\\ &=-\frac{1}{\alpha-a+1}+\frac{1}{\alpha-b+1}\\\\ I(\alpha)&=\int-\frac{1}{\alpha-a+1}+\frac{1}{\alpha-b+1}\:d\alpha=\ln\left| \frac{\alpha-b+1}{\alpha-a+1} \right| + C \end{aligned}$

We still need the integration constant. The integrand uniformly tends to zero as $\alpha\rightarrow-\infty$ :

$\begin{aligned} \lim_{\alpha\rightarrow-\infty} I(\alpha)&\overset{!}{=}0&\Rightarrow&&C=0,&&I(0)&=\ln\left| \frac{1-b}{1-a} \right|\underset{a=2,\:b=4}{=}\ln(\fbox{\$3\$}) \end{aligned}$