Bringing limit under indefinite integral

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Homework Statement



Given:lim_{n\rightarrow ∞} \int^{a^n}_{1} \frac{t^{1/n}}{(1+t)t} dt=\int^{∞}_{1} \frac{1}{(1+t)t} dt

a - Natural number.

I need to prove that I can bring limit under the integral sign.

Homework Equations

The Attempt at a Solution


I've got this so far:
| \int^{a^n}_{1} \frac{t^{1/n}}{(1+t)t} dt-\int^{∞}_{1} \frac{1}{(1+t)t} dt|\stackrel{?}{\rightarrow} 0 while n→∞

| \int^{a^n}_{1} \frac{t^{1/n}}{(1+t)t} dt-\int^{∞}_{1} \frac{1}{(1+t)t} dt|= [did everything I could and wound up with following]=|\int^{a^n}_{1} \frac{t^{1/n}-1}{(1+t)t} dt|

Now I need to either find a function g(t) so f(t)≤g(t) and \int^{a^n}_{1} g(t) dt →0. This is basically the place where I'm stuck and need your help.

p.s. I meant definite integral in the caption.
 
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How do you take the limit as x goes to infinity of an expression made up of a,n and t?
Still, using the fundamental theorem of calculus will help here.
 
Millennial said:
How do you take the limit as x goes to infinity of an expression made up of a,n and t?
Still, using the fundamental theorem of calculus will help here.

Oh, I new I had made some mistakes writing the post. It was meant to be n -> infinity.

And I can't see how can the fundamental theorem of calculus help.
 
In general, let's take the functions f(x,n), a(x,n) and b(x,n), and write this:
\int_{a(x,n)}^{b(x,n)}f(x,n)dx
If F is the antiderivative of f, then we obtain this is equal to
F(b(x,n),n)-F(a(x,n),n)
Now, taking a(x,n)=1, we have
F(b(x,n),n)-F(1,n)

It shouldn't be hard from there.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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