MHF Integral Bee!

**Amer** · August 19th 2009, 12:01 PM

Originally Posted by red_dog

1) Let $f,g:[-a,a]\to\mathbb{R}$ two continuous functions, f odd, g even. Then

$\int_{-a}^a\frac{g(x)}{1+f(x)+\sqrt{1+f^2(x)}}dx=\int_{-a}^af(x)dx$

2) Calculate $\int_{-1}^1\frac{x^{2008}}{1+x^{2007}+\sqrt{1+x^{4014}}}d x$

can I use question one to solve question two without proving one ??

or I should prove one then I can use it

**NonCommAlg** · August 19th 2009, 12:08 PM

Originally Posted by Amer

can I use question one to solve question two without proving one ??

or I should prove one then I can use it

i don't think question 2) needs any solution, since it's just a trivial application of 1).

**red_dog** · August 19th 2009, 12:26 PM

Originally Posted by PaulRS

It should be $ \int_{ - a}^a {\tfrac{{g\left( x \right)}} {{1 + f\left( x \right) + \sqrt {1 + f^2 \left( x \right)} }}dx} = \int_0^a {g\left( x \right) \cdot dx} $

Yes, you're right. Sorry. I've edited.

**NonCommAlg** · August 19th 2009, 01:17 PM

Originally Posted by red_dog

1) Let $f,g:[-a,a]\to\mathbb{R}$ two continuous functions, f odd, g even. Then

$\int_{-a}^a\frac{g(x)}{1+f(x)+\sqrt{1+f^2(x)}}dx=\int_0^a g(x)dx$

ok, let $h(x)=\frac{g(x)}{1+f(x)+\sqrt{1+f^2(x)}}.$ see that $h(x) + h(-x) = g(x).$ also substituting $x \to -x$ gives us $\int_{-a}^0 h(x) \ dx = \int_0^a h(-x) \ dx.$ thus:

$\int_{-a}^a h(x) \ dx = \int_{-a}^0 h(x) \ dx + \int_0^a h(x) \ dx = \int_0^a (h(x) + h(-x)) \ dx = \int_0^a g(x) \ dx.$

Problem: Let $f: [0, \infty) \longrightarrow \mathbb{R}$ be continuous and suppose that $\lim_{x\to\infty} \left(f(x) + \int_0^x f(t) \ dt \right)$ exists and it's finite. Prove that $\lim_{x\to\infty} f(x) = 0.$

**ynj** · August 19th 2009, 06:57 PM

Let $g(x)=\int_0^x f(t) \ dt$
Then $\lim g(x)=\lim \frac{e^x g(x)}{e^x}=\lim\frac{e^x g(x)+e^x g'(x)}{e^x}=\lim g(x)+g'(x)=b$ ,so $\lim f(x)=\lim g'(x)=0$
My problem:
Let $f$ be a polynomial of degree n. $\int_0^1 f(x)x^k \ dx=0, k=1,2,...n$
Prove: $\int_0^1 f^2(x) \ dx=(n+1)^2(\int_0^1 f(x) \ dx)^2$

**simplependulum** · August 19th 2009, 09:24 PM

Let $f(x)$ be $\sum_{k=0}^n a_k x^k$

$\int_0^1 f^2(x)~dx = \int_0^1 f(x)[ \sum_{k=0}^n a_k x^k]~dx$

$= a_0 \int_0^1 f(x)~dx$

And since $\int_0^1 \sum_{k=0}^n a_k x^{k+r}~dx = \sum_{k=0}^n \frac{a_k}{x+r+1} = 0 ~~$ $r=1,2,3...,n$

$\int_0^1 f(x) x^r ~dx$ can be written as

$\frac{k(r-1)(r-2)(r-3)...(r-n)}{ (r+1)(r+2)...(r+n+1)} ~~~~~(1)$

but the numerator is also equal to $\sum_{k=0}^n a_k [\prod_{j\neq k+1}^n (r+j)]~~~(2)$

Then sub $r = 0$ and $r= -1$ into $(1)$ and $(2)$ respectively , obtaining

$\int_0^1 f(x)~dx = \frac{ k (-1)^n}{n+1}$

and

$a_0 (n!) = k (-1)^n (n+1)! = (n+1)! \int_0^1 f(x)~dx (n+1)$

therefore

$\int_0^1 f^2(x)~dx = a_0 \int_0^1 f(x)~dx$

$= (n+1)^2 ( \int_0^1 f(x)~dx)^2$

Problem :

$L^{-1} [ \sqrt{s + \sqrt{s^2 + a^2}} ]$

Is it related to INTEGRAL ? If you think it is not , i still have another problem

**NonCommAlg** · August 19th 2009, 09:57 PM

Originally Posted by ynj

Let $g(x)=\int_0^x f(t) \ dt$
Then $\lim g(x)=\lim \frac{e^x g(x)}{e^x}=\lim\frac{e^x g(x)+e^x g'(x)}{e^x}=\lim g(x)+g'(x)=b$ ,so $\lim f(x)=\lim g'(x)=0$

good thinking but the problem is not that simple!

your proof is not correct because you cannot use L'Hopital unless you know that $\lim_{x\to\infty}e^x g(x) = \infty.$

**ynj** · August 19th 2009, 10:13 PM

Originally Posted by NonCommAlg

good thinking but the problem is not that simple!

your proof is not correct because you cannot use L'Hopital unless you know that $\lim_{x\to\infty}e^x g(x) = \infty.$

not exactly, L'Hospital do not require e^xg(x)->inf,buf just e^x->inf

**ynj** · August 19th 2009, 10:31 PM

maybe you will be confused about what i have said,here is a proof
let $\lim\frac{f'(x)}{g'(x)}=l$ (it exists)
and the only restriction is that $\lim g(x)=\infty,g(x)!=0$
now we will show $\lim\frac{f(x)}{g(x)}=l$
$\forall\epsilon>0,\exists\delta>0,\forall x>\delta,l-\epsilon<\frac{f'(x)}{g'(x)}<l+\epsilon$
so, for some $(c,x)\subset (\delta,\infty), \exists\xi\in (c,x)$
$l-\epsilon<\frac{f(x)-f(c)}{g(x)-g(c)}=\frac{f'(\xi)}{g'(\xi)}<l+\epsilon$
but $\frac{f(x)-f(c)}{g(x)-g(c)}=(\frac{f(x)}{g(x)}-\frac{f(c)}{g(x)})(1-\frac{g(c)}{g(x)})^{-1}$
thus $\frac{f(x)}{g(x)}<\frac{f(c)}{g(x)}+(l+\epsilon)(1-\frac{g(c)}{g(x)})$
let x->inf (sup inf), then
$\limsup\frac{f(x)}{g(x)}\leq l+\epsilon$
let epsilon->0, then
$\limsup\frac{f(x)}{g(x)}\leq l$
similarily,
$\liminf\frac{f(x)}{g(x)}\geq l$
so
$\lim\frac{f(x)}{g(x)}=l$
The proof is copyed from my book. Usually we require numerator to ->inf, but it is unnecessary.

**NonCommAlg** · August 19th 2009, 11:11 PM

Originally Posted by ynj

maybe you will be confused about what i have said,here is a proof
let $\lim\frac{f'(x)}{g'(x)}=l$ (it exists)
and the only restriction is that $\lim g(x)=\infty,g(x)!=0$
now we will show $\lim\frac{f(x)}{g(x)}=l$
$\forall\epsilon>0,\exists\delta>0,\forall x>\delta,l-\epsilon<\frac{f'(x)}{g'(x)}<l+\epsilon$
so, for some $(c,x)\subset (\delta,\infty), \exists\xi\in (c,x)$
$l-\epsilon<\frac{f(x)-f(c)}{g(x)-g(c)}=\frac{f'(\xi)}{g'(\xi)}<l+\epsilon$
but $\frac{f(x)-f(c)}{g(x)-g(c)}=(\frac{f(x)}{g(x)}-\frac{f(c)}{g(x)})(1-\frac{g(c)}{g(x)})^{-1}$
thus $\frac{f(x)}{g(x)}<\frac{f(c)}{g(x)}+(l+\epsilon)(1-\frac{g(c)}{g(x)})$
let x->inf (sup inf), then
$\limsup\frac{f(x)}{g(x)}\leq l+\epsilon$
let epsilon->0, then
$\limsup\frac{f(x)}{g(x)}\leq l$
similarily,
$\liminf\frac{f(x)}{g(x)}\geq l$
so
$\lim\frac{f(x)}{g(x)}=l$
The proof is copyed from my book. Usually we require numerator to ->inf, but it is unnecessary.

i see! you used this theorem! you should've mentioned that in your solution. good work!

**luobo** · August 20th 2009, 07:12 PM

Problem :

$L^{-1} [ \sqrt{s + \sqrt{s^2 + a^2}} ]$

Is it related to INTEGRAL ? If you think it is not , i still have another problem

OK. Let me have a try (first time to do an inverse LaPlace transform). Not sure if it is correct.

Suppose $a>0, Re(s)>0$

$ F(s)\,=\, \sqrt{s + \sqrt{s^2 + a^2}} $

$ 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}} $

$ L^{-1}[F'(s)]\,=\,\frac{a^{\frac{1}{2}}}{2}\cdot J_{-\frac{1}{2}}(at)\cdot u(t) $

$ L^{-1}[F(s)]\,=-\,\frac{a^{\frac{1}{2}}}{2}\cdot J_{-\frac{1}{2}}(at)\cdot \frac{u(t)}{t} $

**luobo** · August 20th 2009, 07:21 PM

Problem: Let $f: [0, \infty) \longrightarrow \mathbb{R}$ be continuous and suppose that $\lim_{x\to\infty} \left(f(x) + \int_0^x f(t) \ dt \right)$ exists and it's finite. Prove that $\lim_{x\to\infty} f(x) = 0.$ [/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".

**simplependulum** · August 20th 2009, 07:33 PM

Originally Posted by luobo

OK. Let me have a try (first time to do an inverse Laplace transform). Not sure if it is correct.

Suppose $a>0, Re(s)>0$

$ F(s)\,=\, \sqrt{s + \sqrt{s^2 + a^2}} $

$ 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}} $

$ L^{-1}[F'(s)]\,=\,\frac{a^{\frac{1}{2}}}{2}\cdot J_{-\frac{1}{2}}(at)\cdot u(t) $

$ L^{-1}[F(s)]\,=-\,\frac{a^{\frac{1}{2}}}{2}\cdot J_{-\frac{1}{2}}(at)\cdot \frac{u(t)}{t} $

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
$J_{-\frac{1}{2}} (at) = \sqrt{\frac{2}{\pi a t}} \cos(at)$
the answer is jsut the function of cosine and t
but what is $u(t)$ ?

**luobo** · August 20th 2009, 07:41 PM

Originally Posted by simplependulum

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
$J_{-\frac{1}{2}} (at) = \sqrt{\frac{2}{\pi a t}} \cos(at)$
the answer is jsut the function of cosine and t
but what is $u(t)$ ?

$u(t)$ is the Heaviside or unit step function, see here:
Heaviside Step Function -- from Wolfram MathWorld

**Amer** · August 21st 2009, 12:16 AM

What is the new integral ?? luobo you should post an integral .

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