A calculus problem by Kushal Bose
Evaluate:
∫ 0 ∞ e x ( ln x ) 2 d x = γ a + c π b
If a , b , c are positive integers then submit your answer as a + b + c
The answer is 10.
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.
1 solution
Wow, this solution looks beast. It looks to me as if you multiplied the integrand by x^(alpha - 1) and differentiated twice w.r.t. alpha to get rid of the (ln(x)^2. But then shouldn't you have a double integral w.r.t. alpha and not a double derivative? Is there something I'm not getting here?
Log in to reply
No. If you start with x α − 1 = x e α ln x and differentiate it twice with respect to α , you obtain x ( ln x ) 2 e α ln x = ( ln x ) 2 x α − 1 . Thus d α 2 d 2 ∫ 0 ∞ e − x x α − 1 d x = ∫ 0 ∞ d α 2 d 2 e − x x α − 1 d x = ∫ 0 ∞ e − x ( ln x ) 2 x α − 1 d x using the necessary theorems to let us bring the double differentiation through the integral sign. Then put α = 1 .
Log in to reply
I figured out where I was mistaken earlier. I wanted to post again but I couldn't find this problem for some reason. Thanks for clarifying anyway. I love seeing your solutions. They are always genius.
We have ∫ 0 ∞ e x ( ln x ) 2 d x = d α 2 d 2 ∫ 0 ∞ e − x x α − 1 d x ∣ ∣ ∣ ∣ α = 1 = Γ ′ ′ ( 1 ) = Γ ( 1 ) [ ψ ( 1 ) 2 + ψ ′ ( 1 ) ] = ψ ( 1 ) 2 + ψ ′ ( 1 ) = γ 2 + 6 1 π 2 making the answer 2 + 2 + 6 = 1 0 .