# Bringing limit under indefinite integral

1. Jun 19, 2012

### PonyBarometer

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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.

Last edited: Jun 19, 2012
2. Jun 19, 2012

### Millennial

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.

3. Jun 19, 2012

### PonyBarometer

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.

4. Jun 19, 2012

### Millennial

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.