ok, let $\displaystyle h(x)=\frac{g(x)}{1+f(x)+\sqrt{1+f^2(x)}}.$ see that $\displaystyle h(x) + h(-x) = g(x).$ also substituting $\displaystyle x \to -x$ gives us $\displaystyle \int_{-a}^0 h(x) \ dx = \int_0^a h(-x) \ dx.$ thus:
$\displaystyle \int_{-a}^a h(x) \ dx = \int_{-a}^0 h(x) \ dx + \int_0^a h(x) \ dx = \int_0^a (h(x) + h(-x)) \ dx = \int_0^a g(x) \ dx.$
Problem: Let $\displaystyle f: [0, \infty) \longrightarrow \mathbb{R}$ be continuous and suppose that $\displaystyle \lim_{x\to\infty} \left(f(x) + \int_0^x f(t) \ dt \right)$ exists and it's finite. Prove that $\displaystyle \lim_{x\to\infty} f(x) = 0.$
Let $\displaystyle g(x)=\int_0^x f(t) \ dt$
Then$\displaystyle \lim g(x)=\lim \frac{e^x g(x)}{e^x}=\lim\frac{e^x g(x)+e^x g'(x)}{e^x}=\lim g(x)+g'(x)=b$,so $\displaystyle \lim f(x)=\lim g'(x)=0$
My problem:
Let$\displaystyle f$be a polynomial of degree n. $\displaystyle \int_0^1 f(x)x^k \ dx=0, k=1,2,...n$
Prove:$\displaystyle \int_0^1 f^2(x) \ dx=(n+1)^2(\int_0^1 f(x) \ dx)^2$
Let $\displaystyle f(x)$ be $\displaystyle \sum_{k=0}^n a_k x^k $
$\displaystyle \int_0^1 f^2(x)~dx = \int_0^1 f(x)[ \sum_{k=0}^n a_k x^k]~dx$
$\displaystyle = a_0 \int_0^1 f(x)~dx $
And since $\displaystyle \int_0^1 \sum_{k=0}^n a_k x^{k+r}~dx = \sum_{k=0}^n \frac{a_k}{x+r+1} = 0 ~~ $ $\displaystyle r=1,2,3...,n $
$\displaystyle \int_0^1 f(x) x^r ~dx $ can be written as
$\displaystyle \frac{k(r-1)(r-2)(r-3)...(r-n)}{ (r+1)(r+2)...(r+n+1)} ~~~~~(1)$
but the numerator is also equal to $\displaystyle \sum_{k=0}^n a_k [\prod_{j\neq k+1}^n (r+j)]~~~(2)$
Then sub $\displaystyle r = 0 $ and $\displaystyle r= -1$ into $\displaystyle (1) $ and $\displaystyle (2) $ respectively , obtaining
$\displaystyle \int_0^1 f(x)~dx = \frac{ k (-1)^n}{n+1} $
and
$\displaystyle a_0 (n!) = k (-1)^n (n+1)! = (n+1)! \int_0^1 f(x)~dx (n+1)$
therefore
$\displaystyle \int_0^1 f^2(x)~dx = a_0 \int_0^1 f(x)~dx $
$\displaystyle = (n+1)^2 ( \int_0^1 f(x)~dx)^2$
Problem :
$\displaystyle L^{-1} [ \sqrt{s + \sqrt{s^2 + a^2}} ] $
Is it related to INTEGRAL ? If you think it is not , i still have another problem
maybe you will be confused about what i have said,here is a proof
let $\displaystyle \lim\frac{f'(x)}{g'(x)}=l$(it exists)
and the only restriction is that $\displaystyle \lim g(x)=\infty,g(x)!=0$
now we will show $\displaystyle \lim\frac{f(x)}{g(x)}=l$
$\displaystyle \forall\epsilon>0,\exists\delta>0,\forall x>\delta,l-\epsilon<\frac{f'(x)}{g'(x)}<l+\epsilon$
so, for some$\displaystyle (c,x)\subset (\delta,\infty), \exists\xi\in (c,x)$
$\displaystyle l-\epsilon<\frac{f(x)-f(c)}{g(x)-g(c)}=\frac{f'(\xi)}{g'(\xi)}<l+\epsilon$
but $\displaystyle \frac{f(x)-f(c)}{g(x)-g(c)}=(\frac{f(x)}{g(x)}-\frac{f(c)}{g(x)})(1-\frac{g(c)}{g(x)})^{-1}$
thus$\displaystyle \frac{f(x)}{g(x)}<\frac{f(c)}{g(x)}+(l+\epsilon)(1-\frac{g(c)}{g(x)})$
let x->inf (sup inf), then
$\displaystyle \limsup\frac{f(x)}{g(x)}\leq l+\epsilon$
let epsilon->0, then
$\displaystyle \limsup\frac{f(x)}{g(x)}\leq l$
similarily,
$\displaystyle \liminf\frac{f(x)}{g(x)}\geq l$
so
$\displaystyle \lim\frac{f(x)}{g(x)}=l$
The proof is copyed from my book. Usually we require numerator to ->inf, but it is unnecessary.
OK. Let me have a try (first time to do an inverse LaPlace transform). Not sure if it is correct.Problem :
$\displaystyle L^{-1} [ \sqrt{s + \sqrt{s^2 + a^2}} ] $
Is it related to INTEGRAL ? If you think it is not , i still have another problem
Suppose $\displaystyle a>0, Re(s)>0 $
$\displaystyle
F(s)\,=\, \sqrt{s + \sqrt{s^2 + a^2}}
$
$\displaystyle
F'(s)\,=\, \frac{\sqrt{s+\sqrt{s^2+a^2}}}{2\sqrt{s^2+a^2}}\,= \, \frac{a^{\frac{1}{2}}}{2}\cdot\frac{a^{-\frac{1}{2}}(s+\sqrt{s^2+a^2})^{\frac{1}{2}}}{\sqr t{s^2+a^2}}
$
$\displaystyle
L^{-1}[F'(s)]\,=\,\frac{a^{\frac{1}{2}}}{2}\cdot J_{-\frac{1}{2}}(at)\cdot u(t)
$
$\displaystyle
L^{-1}[F(s)]\,=-\,\frac{a^{\frac{1}{2}}}{2}\cdot J_{-\frac{1}{2}}(at)\cdot \frac{u(t)}{t}
$
Problem: Let $\displaystyle f: [0, \infty) \longrightarrow \mathbb{R}$ be continuous and suppose that $\displaystyle \lim_{x\to\infty} \left(f(x) + \int_0^x f(t) \ dt \right)$ exists and it's finite. Prove that $\displaystyle \lim_{x\to\infty} f(x) = 0.$[/quote]
The solution is here.
l'Hôpital's rule - Wikipedia, the free encyclopedia
http://upload.wikimedia.org/math/5/f...c513fb6556.png
Check last paragraph in Section "Examples".
wow, I don't know the inverse Laplace Transform can be written in terms of Bessel Fucntion I have another solution but your answer should be correct ,
since
$\displaystyle J_{-\frac{1}{2}} (at) = \sqrt{\frac{2}{\pi a t}} \cos(at) $
the answer is jsut the function of cosine and t
but what is $\displaystyle u(t)$ ?
$\displaystyle u(t) $ is the Heaviside or unit step function, see here:
Heaviside Step Function -- from Wolfram MathWorld