lebesgue integration

alexthepenguin · #1 $Old$ November 25th, 2008, 02:23 AM

i have exams on this course coming up in two days (i should probably write this post earlier). i was hoping to read more and try to find a solution myself but unfortunately failed.
let f be a positive function, integrable over R. prove that

$\sum_{n \in Z} \int_0^1 f(x+n) d \mu = \int_{-\infty}^{\infty} f(x) d \mu$
and deduce that $\sum_{n \in Z} f(x+n)$ converges

prove that

$\lim_{n\rightarrow \infty} \sqrt{n} \int_0^{\pi/2} \cos^n x d\mu=\int_0^\infty e^{-x^2/2}$

this is based on the assumption that $\ln(\cos x) \le \frac{-x^2}{2}$ for all x in 0..pi/2 ( i have done this part)

also one mroe if p >-1 and q belongs to the natural number. prove that

$\lim_{n\rightarrow \infty} \int_0^n x^p(ln x)^q(1-x/n)^n d\mu=\lim_{n\rightarrow \infty} \int_0^\infty x^p (ln x)^q e^-x d\mu$ (i have done this part)

deduce that
$\int_0^\infty e^-x ln x d\mu =\lim_{n\rightarrow \infty}(ln x-(1+1/2+...+1/n))$

any thoughts would be appreciated! thank you!

David24 · #2 $Old$ November 25th, 2008, 03:17 AM

Quote:

Originally Posted by alexthepenguin $View Post$

i have exams on this course coming up in two days (i should probably write this post earlier). i was hoping to read more and try to find a solution myself but unfortunately failed.
let f be a positive function, integrable over R. prove that

$\sum_{n \in Z} \int_0^1 f(x+n) d \mu = \int_{-\infty}^{\infty} f(x) d \mu$
and deduce that $\sum_{n \in Z} f(x+n)$ converges

prove that

$\lim_{n\rightarrow \infty} \sqrt{n} \int_0^{\pi/2} \cos^n x d\mu=\int_0^\infty e^{-x^2/2}$

this is based on the assumption that $\ln(\cos x) \le \frac{-x^2}{2}$ for all x in 0..pi/2 ( i have done this part)

also one mroe if p >-1 and q belongs to the natural number. prove that

$\lim_{n\rightarrow \infty} \int_0^n x^p(ln x)^q(1-x/n)^n d\mu=\lim_{n\rightarrow \infty} \int_0^\infty x^p (ln x)^q e^-x d\mu$ (i have done this part)

deduce that
$\int_0^\infty e^-x ln x d\mu =\lim_{n\rightarrow \infty}(ln x-(1+1/2+...+1/n))$

any thoughts would be appreciated! thank you!

Okay,

Ive done the first one,

LHS = sum ( int(f(x+n) x = 0 to 1), n = -inf, ..., inf)

consider the integral component,

inf(f(x+n) , x = 0 to 1)
here let t = x + n --> dt/dx = 1 or dt = dx
Change Limits: at x = 0 t = n, at x = 1 t = n +1

Thus,

inf(f(x+n), x = 0 to 1) becomes int( f(t) t = n to n+1) or int(f(x), x = n to n+1)

Plug back into the sum,

sum ( int(f(x), x = n to n + 1) , n = -inf,...,inf)

which as Im sure you can easily see is
sum(int(f(x), x = -inf...inf))

Hope this helps,

I'll keep plugging away on the rest,

Regards,

David

Opalg · #3 $Old$ November 25th, 2008, 05:53 AM

Quote:

Originally Posted by alexthepenguin $View Post$

prove that

$\lim_{n\rightarrow \infty} \sqrt{n} \int_0^{\pi/2} \cos^n x d\mu=\int_0^\infty e^{-x^2/2}$

this is based on the assumption that $\ln(\cos x) \le \frac{-x^2}{2}$ for all x in 0..pi/2 ( i have done this part)

Substitute y = x√n to get $\sqrt n\int_0^{\pi/2}\cos^nx\,dx = \int_0^{\pi\sqrt n/2}\cos^n(y/\sqrt n)\,dy$ .

Use good ol' l'Hôpital to check that $\lim_{x\to0}\frac{\ln(\cos x)}{x^2} = \frac12$ . Put $x=y/\sqrt n$ in that, to see that $\lim_{n\to\infty}n\ln\bigl(\cos(y/\sqrt n)\bigr) = -\frac{y^2}2$ . Taking exponentials, $\lim_{n\to\infty}\cos^n(y/\sqrt n) = e^{-y^2/2}$ . Also, the inequality $\ln(\cos x)\leqslant-x^2/2$ shows that $\cos^n(y/\sqrt n) \leqslant e^{-y^2/2}$ . You can then apply the Dominated Convergence Theorem to get $\lim_{n\to\infty} \int_0^{\pi\sqrt n/2}\cos^n(y/\sqrt n)\,dy = \int_0^\infty e^{-y^2/2}dy$ .