A = ∫ 1 ∞ ( u 2 1 − u 4 1 ) ln u d u
Find the value of e A .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Bravo, Sir! I would've never thought to do such a trick. It works so smoothly. Thanks for contributing this problem.
This is an instance of Frullani integral in disguise. Substitute u = e x . Then A = ∫ 0 ∞ x e − x − e − 3 x d x , and so, the answer is lo g 3 . For those who are not familiar with Frullani integral, the trick is to write x e − x − e − 3 x = ∫ 1 3 e − s x 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 s d s = lo g 3 .
Put ln u = t, and use the Frullani Integral.
Another pedestrian approach with Differentiation Under the Integral Sign....
Let u = e − x
A = ∫ 0 − ∞ ( e − 2 x 1 − e − 4 x 1 ) − x − e − x d x
Rearranging....
A = ∫ − ∞ 0 x e 3 x − x e x d x
Let I ( n ) = ∫ − ∞ 0 x e n x d x so that A = I ( 3 ) − I ( 1 )
Taking the derivative of I ( n ) with respect to n yields
I ′ ( n ) = ∫ − ∞ 0 ∂ n ∂ x e n x d x = ∫ − ∞ 0 e n x d x = n 1
And re-integrating leads to
I ( n ) = ∫ I ′ ( n ) d n = ∫ n 1 d n = l n ( n ) + C
And thus
A = I ( 3 ) − I ( 1 ) = l n ( 3 ) + C − l n ( 1 ) − C = l n ( 3 )
Therefore e A = e l n ( 3 ) = 3
Another possibility to apply Differentiation Under the Integral Sign :
I ( α ) : = ∫ 1 ∞ u α ( u − a − u − b ) ln ( u ) d u a , b > 1 , α < min { a ; b } − 1
We want to calculate I ( 0 ) and start with the derivative of I ( α ) :
I ( 1 ) ( α ) I ( α ) = ∫ 1 ∞ u α ( u − a − u − b ) d u = [ α − a + 1 u α − a + 1 − α − b + 1 u α − b + 1 ] 1 ∞ = − α − a + 1 1 + α − b + 1 1 = ∫ − α − a + 1 1 + α − b + 1 1 d α = ln ∣ ∣ ∣ ∣ α − a + 1 α − b + 1 ∣ ∣ ∣ ∣ + C
We still need the integration constant. The integrand uniformly tends to zero as α → − ∞ :
α → − ∞ lim I ( α ) = ! 0 ⇒ C = 0 , I ( 0 ) = ln ∣ ∣ ∣ ∣ 1 − a 1 − b ∣ ∣ ∣ ∣ a = 2 , b = 4 = ln ( $ 3 $ )
Problem Loading...
Note Loading...
Set Loading...
Relevant wiki: Differentiation Under the Integral Sign
Let a , b ∈ R with a , b > 1 , and let f ( a , b ) = ∫ 1 ∞ ( x a 1 − x b 1 ) ln ( x ) d x . Then,
d b d ( f ( a , b ) ) = = = = d b d ( ∫ 1 ∞ ( x a 1 − x b 1 ) ln ( x ) d x ) ∫ 1 ∞ d b d ( x a 1 − x b 1 ) ln ( x ) d x ∫ 1 ∞ x b 1 d x b − 1 1
Therefore, f ( a , b ) = ln ∣ b − 1 ∣ + C
Since f ( a , a ) = 0 , then C = − ln ∣ a − 1 ∣ . Hence, f ( a , b ) = ∫ 1 ∞ ( x a 1 − x b 1 ) ln ( x ) d x = ln ∣ ∣ ∣ ∣ a − 1 b − 1 ∣ ∣ ∣ ∣
Therefore, e f ( 2 , 4 ) = e ln ( 3 ) = 3 .