int question (1 Viewer)

[sikeveo]

lim(n-> ∞) ∫ sin(x/n) / (1 + x/n)^n dx
where the integral is between 0 and ∞.

How would you go about solving this?

 

[robbo_145]

interesting question..
my 2 cents.
substitution

let u = x/n
ndu = dx
thus I becomes
I = lim(n-> ∞) n∫ sin(u) / (1 + u)^n du
from 0 to 1 (assuming the 'infinite' you are going to is one and the same)

as n gets large the integral obviously tends to 0 as sinu < u < 1 + u << (1+u)^n
thus I = 0
i dunno how valid this is... but hey im first year . i eagerly await other replies

edit: cant spell

 

[sikeveo]

Hmm, yeh i did something similar but it doesn't look right...

 

[hyparzero]

let u = x/n
ndu = dx
thus I becomes
I = lim(n-> ∞) n∫ sin(u) / (1 + u)^n du

Using reduction....

=> I = lim(n-∞) n[ - (sinu)/(n-1)(1+x)^n-1 + [1/(n-1)] * [∫ cosu/(1+x)^n-1 ]du

Definite Function is still between 0 and ∞ for u = x/n

As n -> ∞, we see that - (sinu)/(n-1)(1+x)^n-1 = 0

and similarly [1/(n-1)] = 0

Therefore I = ∞( 0 + 0) = Undefined

This simply means that for the function sin(x/n) / (1 + x/n)^n, it diverges between x=0 to ∞ for all values of n.

 

[Yip]

I tend to agree with robbo_145's solution. If you just graph the function being integrated (via mathematica or similar sofware[i kno im cheap ]), its clear that on the interval [0,infinity], the area enclosed by the curve and x-axis tends towards 0 as n increases