# Let g(x)=$\frac1x\int_1^xf(t)dt$; x>0, and assume $\lim_{x\to \infty}g(x)= B$ exists. Show the following.

Let f : [0,$$\infty$$) -> $$\mathbb R$$ be a continuous function and let g(x)=$$\frac1x\int_1^xf(t)dt$$; x>0.

Assume that $$\lim_{x\to \infty}g(x)= B$$ exists. Let 0 < a < b be two fixed numbers.

Show:

(i) $$\int_a^b\frac{f(x)}{x}dx$$ = g(b) - g(a) + $$\int_a^b\frac{g(x)}{x}dx$$ (Hint: Use Integration by Parts.)

(ii) $$\lim_{T\to \infty}\int_{Ta}^{Tb}\frac{f(x)}{x}dx=Bln(\frac ba)$$

I tried integral by parts for part (i) but I'm not sure how it would work with an unknown function f(x). It gave me:

$$\left[\frac 1x \int f(x)\right]_a^b + \int_a^b\left(\frac 1{x^2}\int f(x)\right)dx$$

which does kind of look like what we're looking for, except for the limits of the integral of f(x), I don't know how to work with that, or how it's supposed to go from that to [1,x], as needed in g(x).

Then for (ii) just plug in what we got for (i), and using the assumption that $$\lim_{x\to \infty}g(x)= B$$, and then substituting the limits, I get:

$$\lim_{T\to \infty}\int_{a}^{b}\frac{g(Tx)}{x}dx$$

And not sure what to do now... Considered IBP again, but wouldn't know how to do that with g(Tx). Please let me know what you think

For (i), $$g(x)=\frac1x\int_1^xf(t)dt$$ $$xg(x)=\int_1^xf(t)dt$$ Take derivative with respect to $$x$$ on both sides, $$x\frac{dg(x)}{dx}+g(x)=f(x)$$ Put value of $$f(x)$$ to get, $$\int_a^b\frac{f(x)}{x}dx=\int_a^b\frac{dg(x)}{dx}dx+\int_a^b\frac{g(x)}{x}dx$$ $$=\int_a^bdg(x)+\int_a^b\frac{g(x)}{x}dx$$ $$=[g(x)]_a^b+\int_a^b\frac{g(x)}{x}dx$$ $$=g(b)-g(a)+\int_a^b\frac{g(x)}{x}dx$$
For (ii), $$\lim_{T\to \infty}\int_{Ta}^{Tb}\frac{f(x)}{x}dx=\lim_{T\to \infty}\left(g(Tb)-g(Ta)+\int_{Ta}^{Tb}\frac{g(x)}{x}dx \right)$$ $$=\lim_{T\to \infty}g(Tb)-\lim_{T\to \infty}g(Ta)+\lim_{T\to \infty}\left(\int_{Ta}^{Tb}\frac{g(x)}{x}dx \right)$$ $$=\lim_{T\to \infty}\left(\int_{Ta}^{Tb}\frac{g(x)}{x}dx \right)$$ Put $$z=\frac{x}{T}$$ so, $$dz=\frac{dx}{T}$$, $$=\lim_{T\to \infty}\left(\int_a^b\frac{g(Tz)}{Tz}Tdz \right)$$ $$=\int_a^b\lim_{T\to \infty}\left(\frac{g(Tz)}{z}dz \right)$$ $$=\int_a^b\lim_{T\to \infty}g(Tz)\lim_{T\to \infty}\left(\frac{dz}{z} \right)$$ But $$\lim_{T\to \infty}g(Tz)=B$$ and $$\lim_{T\to \infty}\left(\frac{dz}{z} \right)=\left(\frac{dz}{z} \right)$$ $$=B\int_a^b\frac{dz}{z}$$ $$=Blog \left(\frac{b}{a} \right)$$
Let $$h$$ be an antiderivative for $$f$$. Then $$\int_a^{b} \frac {f(t)} t\, dt= \frac 1 t h(t)|_a^{b}-\int_a^{b} (-\frac 1 {t^{2}}) h(t)\, dt$$. Since $$tg(t)$$ is an antiderivative for $$f$$ we can put $$h(t)=tg(t)$$. Hence $$\int_a^{b} \frac {f(t)} t\, dt= g(b)-g(a)+\int_a^{b} \frac {g(t)} t\, dt$$.
Now $$\int_{Ta}^{Tb} \frac {f(t)} t\, dt= g(Tb)-g(Ta)+\int_{Ta}^{Tb} \frac {g(t)} t\, dt$$. Use squeeze theorem to show that the limit of the last term is same as limit of $$\int_{Ta}^{Tb} \frac {B} t\, dt$$ which is limit of $$B(\log\, (Tb)-\log\, (Tb))=B\log \, (\frac b a)$$.
For (i), let $$F(x) = \int_1^xf(t)\,dt$$, so $$F'(x) = f(x)$$ by the fundamental theorem of calculus. Then our integration by parts looks like $$\int_a^b\frac{f(x)}{x}\,dx = \bigg[\frac{F(x)}{x}\bigg]_a^b + \int_a^b \frac{F(x)}{x^2}\,dx.$$
For (ii), using our result from part (i), $$\int_{Ta}^{Tb}\frac{f(x)}{x}\,dx = \bigg[\frac{F(x)}{x}\bigg]_{Ta}^{Tb} + \int_{Ta}^{Tb} \frac{F(x)}{x^2}\,dx = \big[g(x)\big]_{Ta}^{Tb} + \int_{Ta}^{Tb} \frac{g(x)}{x}\,dx$$ For sufficiently large $$T$$, show that the first term can be made arbitrarily small, and the second term is approximately $$\int_{Ta}^{Tb}B/x\,dx = B\ln(b/a)$$.