Logarithmic Integrals?

Calculus Level 2

A = 1 ( 1 u 2 1 u 4 ) d u ln u \large A = \displaystyle\int_{1}^{\infty} \left(\dfrac{1}{u^2} - \dfrac{1}{u^4}\right) \dfrac{du}{\ln u}

Find the value of e A e^A .


The answer is 3.

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5 solutions

Ariel Gershon
Sep 15, 2016

Relevant wiki: Differentiation Under the Integral Sign

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

d ( f ( a , b ) ) d b = d d b ( 1 ( 1 x a 1 x b ) d x ln ( x ) ) = 1 d d b ( 1 x a 1 x b ) d x ln ( x ) = 1 1 x b d x = 1 b 1 \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 f(a,b) = \ln |b-1| + C

Since f ( a , a ) = 0 f(a,a) = 0 , then C = ln a 1 C = -\ln |a-1| . Hence, f ( a , b ) = 1 ( 1 x a 1 x b ) d x ln ( x ) = ln b 1 a 1 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 ) = 3 e^{f(2,4)} = e^{\ln(3)} = \boxed{3} .

Bravo, Sir! I would've never thought to do such a trick. It works so smoothly. Thanks for contributing this problem.

James Wilson - 3 years, 10 months ago
Sangchul Lee
Mar 19, 2019

This is an instance of Frullani integral in disguise. Substitute u = e x u = e^x . Then A = 0 e x e 3 x x d x , A = \int_{0}^{\infty} \frac{e^{-x} - e^{-3x}}{x} \, \mathrm{d}x, and so, the answer is log 3 \log 3 . For those who are not familiar with Frullani integral, the trick is to write e x e 3 x x = 1 3 e s x d s \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 = 0 ( 1 3 e s x d s ) d x = 1 3 ( 0 e s x d x ) d s = 1 3 d s s = log 3. 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.

Les Schumer
May 18, 2020

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

Let u = e x u = e^{-x}

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

Rearranging....

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

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

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

I ( n ) = 0 n e n x x d x = 0 e n x d x = 1 n 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 ) = I ( n ) d n = 1 n d n = l n ( n ) + C I(n) = \int I'(n) dn = \int \frac{1}{n} dn = ln(n) + C

And thus

A = I ( 3 ) I ( 1 ) = l n ( 3 ) + C l n ( 1 ) C = l n ( 3 ) A = I(3) - I(1) = ln(3) + C - ln(1) - C = ln(3)

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

Carsten Meyer
Feb 20, 2019

Another possibility to apply Differentiation Under the Integral Sign :

I ( α ) : = 1 u α ( u a u b ) d u ln ( u ) a , b > 1 , α < min { a ; b } 1 \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 ) I(0) and start with the derivative of I ( α ) I(\alpha) :

I ( 1 ) ( α ) = 1 u α ( u a u b ) d u = [ u α a + 1 α a + 1 u α b + 1 α b + 1 ] 1 = 1 α a + 1 + 1 α b + 1 I ( α ) = 1 α a + 1 + 1 α b + 1 d α = ln α b + 1 α a + 1 + C \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 α \alpha\rightarrow-\infty :

lim α I ( α ) = ! 0 C = 0 , I ( 0 ) = ln 1 b 1 a = a = 2 , b = 4 ln ( $3$ ) \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}

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