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Science Forum Index » Mathematics Forum » Does this integral converges?
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| Carl R. |
 Posted: Sat May 03, 2008 5:26 pm |
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Hello,
Let f(x) = 1/sqrt(x + x^2). Is the following integral convergent?
Int ( f(x) dx, x, 0, infinity).
it's easy to see that 1/sqrt(x + x^2 ) < 1/ sqrt(x) but that doesn't
really help ( I think) because the right hand side doesn't converge,
which inequality would be useful for this particular problem?
I think I should find a function g(x) such that f(x) > g(x) and g(x)
diverges in (0,infinity) but haven't had success.
Thanks |
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| Arturo Magidin |
| Posted: Sat May 03, 2008 5:34 pm |
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In article <15e0e5f8-9ad8-4719-8de2-02d9e6c00fed@t12g2000prg.googlegroups.com>,
Carl R. <solrac140@hotmail.com> wrote:
Quote:
Hello,
Let f(x) = 1/sqrt(x + x^2). Is the following integral convergent?
Int ( f(x) dx, x, 0, infinity).
it's easy to see that 1/sqrt(x + x^2 ) < 1/ sqrt(x) but that doesn't
really help ( I think) because the right hand side doesn't converge,
which inequality would be useful for this particular problem?
Try
x + x^2 < x^2 + x^2 = 2x^2 (which holds for x>1).
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org |
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| Carl R. |
| Posted: Sat May 03, 2008 5:47 pm |
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On 3 mayo, 20:34, magi...@math.berkeley.edu (Arturo Magidin) wrote:
Quote: In article <15e0e5f8-9ad8-4719-8de2-02d9e6c00...@t12g2000prg.googlegroups.com>,
Carl R. <solrac...@hotmail.com> wrote:
Hello,
Let f(x) = 1/sqrt(x + x^2). Is the following integral convergent?
Int ( f(x) dx, x, 0, infinity).
it's easy to see that 1/sqrt(x + x^2 ) < 1/ sqrt(x) but that doesn't
really help ( I think) because the right hand side doesn't converge,
which inequality would be useful for this particular problem?
Try
x + x^2 < x^2 + x^2 = 2x^2 (which holds for x>1).
--
=====================================================================> "It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
=====================================================================
Arturo Magidin
magidin-at-member-ams-org
Oh that's nice, thanks, your idea is to split the integral from 0 to 1
and from 1 to infinity
and then compare it to 2*x^2, then the integral of f from 1 to
infinity diverges.
My question is, when splitting the improper integral as the sum of two
integrals, if one of them diverges then the whole integral diverges?
this comes from the definition, right? or why it suffices to look at
only one part of the whole interval?
Thanks |
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| Arturo Magidin |
| Posted: Sat May 03, 2008 6:14 pm |
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In article <9460a3ae-3c78-414e-94c1-338e57a81e4c@q27g2000prf.googlegroups.com>,
Carl R. <solrac140@hotmail.com> wrote:
Quote: On 3 mayo, 20:34, magi...@math.berkeley.edu (Arturo Magidin) wrote:
In article <15e0e5f8-9ad8-4719-8de2-02d9e6c00...@t12g2000prg.googlegroups.=
com>,
Carl R. <solrac...@hotmail.com> wrote:
Hello,
Let f(x) =3D 1/sqrt(x + x^2). Is the following integral convergent?
Int ( f(x) dx, x, 0, infinity).
it's easy to see that 1/sqrt(x + x^2 ) < 1/ sqrt(x) but that doesn't
really help ( I think) because the right hand side doesn't converge,
which inequality would be useful for this particular problem?
Try
x + x^2 < x^2 + x^2 =3D 2x^2 =A0(which holds for x>1).
Oh that's nice, thanks, your idea is to split the integral from 0 to 1
and from 1 to infinity
and then compare it to 2*x^2, then the integral of f from 1 to
infinity diverges.
You do realize that your integral is improper on BOTH ends, right?
Both at 0 and at oo. So in ->any<- case, you need to pick a value c,
0< c < oo, and divide the integral; the interval from 0 to oo will
converge if and only if the integral from 0 to c and the integral from
c to oo BOTH converge. That's part of the definition of improper
integral. It can be shown that whether or not it converges is
independent of the choice of c; see below.
One good place to break it up is at c=1. From the integral from 1 to
oo, you can do the comparison I pointed out, and at that point you are
done (since the integral will diverge there).
Quote: My question is, when splitting the improper integral as the sum of two
integrals, if one of them diverges then the whole integral diverges?
this comes from the definition, right? or why it suffices to look at
only one part of the whole interval?
Think about the definition of the integral from a to infinity. It is
really a limit: the limit, as b->oo, of the integral from a to b.
The integral from a to b is equal to the integral from a to c plus the
integral from c to b (provided the function is defined on an interval
that contains both a, b, and c). And the limit as b->oo of the
integral from a to c is just the integral, since it does not depend on
b. There is a theorem from limits that says that the limit of a+f(x)
(x being the variable) exists if and only if the limit of f(x) exists,
so you are applying that.
For an integral that has "problems on both ends" (like the integral
you have here, from 0 to oo), if c and d are any positive numbers, the
integral from 0 to oo is the same as either the sum of the integral
from 0 to c plus the integral from c to infinity (and exists if and
only if both of these exist, in whch case it equals the sum of the
two), and also equal to the sum of the integrals from 0 to d plus the
integral from d to infinity (and exists if and only if both of them
exist, in which case it equals the sum of the two). If both exist when
you split them at c, you have
integral (0 to oo) f(x) dx
= integral(0 to c)f(x) dx + integral(c to oo) f(x)dx
= lim(a->0+) integral(a to c)f(x)dx
+ lim(b->oo) integral(c to b)f(x)dx
= limt(a->0+) integral(a to c)f(x)dx
+ lim(b->oo)[integral(c to d)f(x)dx + integral(d to b)f(x)dx
= limt(a->0+) integral(a to c)f(x)dx + integral(c to d)f(x)dx
+ lim(b->oo) integral(d to b) f(x)dx
= lim(a->0+)[integral(a to c)f(x)dx + integral(c to d)f(x)dx]
+ lim(b->oo) integral(d to b)f(x)dx
= lim(a->0+)integral(a to d)f(x)dx
+ lim(b->oo) integral(d to b)f(x)dx
= integral(0 to d)f(x)dx + integral(d to oo)f(x)dx
so you get the same answer no matter how you split it (if the limits
exists). It is easy to check that if integral(c to oo)f(x)dx diverges
then so does integral(d to oo)f(x)dx (again, just split the integral
inside the limit), and likewise that integral(0 to c)f(x)dx converges
if and only if integral(0 to d)f(x)dx converges (this is assuming the
ONLY problems are at the endpoints, of course; again, split the
"longer" integral inside the limit). So if any of the limits do not
exist, then they do not exist no matter how you split the integral. In
summary, you need to split the integral, and it doesn't matter how you
do it, so long as you do it.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org |
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| William Elliot |
| Posted: Sat May 03, 2008 11:01 pm |
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On Sat, 3 May 2008, Carl R. wrote:
Quote: Let f(x) = 1/sqrt(x + x^2). Is the following integral convergent?
u = sqr x; sqr(1 + x) = sqr(1 + u^2)
Quote: Int ( f(x) dx, x, 0, infinity).
integral(0,oo) f(x) dx = 2.integral(0,oo) 1/sqr(1 + u^2) du
Quote: it's easy to see that 1/sqrt(x + x^2 ) < 1/ sqrt(x) but that doesn't
really help ( I think) because the right hand side doesn't converge,
which inequality would be useful for this particular problem?
I think I should find a function g(x) such that f(x) > g(x) and g(x)
diverges in (0,infinity) but haven't had success.
Thanks
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