Infinite graph with finite area?

[kreil]

Is it possible for the graph of a function to extend to infinity in either direction but have a finite area under the curve?

It seems to me that since series exist that have infinite terms but are convergent, the same holds true here.

If so, does anyone have an example function I could play around with?

Thanks

 

[HallsofIvy]

Yes, of course. The function would have to go to 0, rapidly, in both directions. One important example is the "error function",
\frac{1}{2\pi}e^{-x^2}
which has area under the curve of 1.

 

[dextercioby]

How about e^{-kx} [/tex] with &#039;k&quot; and &quot;x&quot; strictly positive,equivalently,&quot;k&quot; and &quot;x&quot; strictly negative ?<br /> <br /> Daniel.

 

[matt grime]

How about e^{-kx} [/tex] with &#039;k&quot; and &quot;x&quot; strictly positive,equivalently,&quot;k&quot; and &quot;x&quot; strictly negative ?<br /> <br /> Daniel.

<br /> <br /> <br /> |kx| you mean?

 

[dextercioby]

If you prefer,i was referring to this integral

\int_{0}^{+\infty} e^{-kx} dx

,k>0.

Then it occurred to me that
\int_{-\infty}^{0} e^{-kx} dx

k<0

is the same thing...

Daniel.

P.S.I don't like modulus under the integral sign...

 

[Galileo]

\int \limits_{-\infty}^{\infty}\frac{dx}{1+x^2}

Easily integrated with a beautiful result.
This is one of my favourite integrals... (boy, that sounded nerdish).

 

[kreil]

Thanks for all the equations

Would someone take the time to show me the algebra for

\int_{-\infty}^{\infty}(\frac{1}{2\pi}e^{-x^2})dx=1

I don't know how to solve integrals at infinity, the only thing I know how to use to evaluate integrals right now is the fundamental theorem (I'm a high school senior in AP calc AB, but I plan to be a math major).

 

[dextercioby]

OMG,i saw it just now...HALLS,what did u do?? Where is the square root?And where did that 2 come from??

Daniel.

P.S.O.P.1-st question:Do you know Fubini's theorem??If u don't,are u willing to accept that the product of 2 improper integrals on R is equivalent with a double improper integral on R^{2}??

 

[Galileo]

Integrals at infinity are defined by a limit, just like sequences or series.
Similarly, if the integral limit exists, the integral is called convergent.

\int_a^{\infty}f(x)dx=\lim_{t \to \infty} \int_a^t f(x)dx.

For \int \limits_{-\infty}^{\infty}\exp(-x^2)dx you need a little trick which uses a double integral.

 

[matt grime]

OMG,i saw it just now...HALLS,what did u do?? Where is the square root?And where did that 2 come from??

Daniel.

P.S.O.P.1-st question:Do you know Fubini's theorem??If u don't,are u willing to accept that the product of 2 improper integrals on R is equivalent with a double improper integral on R^{2}??

come on! 2pi=1, so root(2pi)=2pi

The OP says he only knows the fundamental theorem of calc so I'm guessing fubini's theorem is not on the syllabus.

For the OP. It can't be done by the elementary means he knows, since it isn't an elementary integral, ie the integrand isn't the derivative of something nice.

 

[dextercioby]

Okay,Matt,how about the DEFINITION OF ERF? Or is that the trully "relative" thing...??

Daniel.

EDIT:That should be actually:

2\pi\cdot n\cdot i =1 ;n\in Z^{*}

 

[Curious3141]

come on! 2pi=1, so root(2pi)=2pi

Isn't it actually 2\pi i[/tex] that&#039;s actually 1 ? <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /><br /> <br /> EDIT : it&#039;s actually zero, d&#039;oh !

 

[matt grime]

If 2pi were 1 then no one would make a mistake in Fourier series again... apart from the other mistakes that is.

 

[HallsofIvy]

come on! 2pi=1, so root(2pi)=2pi

I'll have to remember that! Yes, I dropped the square root.

kreil: to integrat \int_{-\infty}^\infty e^{-x^2}dx, first write
I= \int_{-\inty}^\infty e^{-x^2}dx. Because of the symmetry,
\frac{I}{2}= \int_0^\infty e^{-x^2}dx. It is also true that \frac{I}{2}= \int_0^\infty e^{-y^2}dy since that's just a change in dummy variable.

Multiply the two together: \frac{I^2}{4}= \int_0^\infty e^{-x^2}dx \int_0^\infty e^{-x^2}dx. Fubini's theorm, that dextercioby referred to, basically says that that product is the same as the double or repeated integral (which Fubini's theorem also says are the same) \int_{x=0}^{\infty}\int_{y=0}^{\infty}e^{-(x^2+ y^2)} dydx. Think of that as an integral over the first quadrant in the xy-plane and convert to polar coordinates: x²+ y[/sup]2[/sup] becomes r² and dydx become r dr dθ. See that "r" in the differential? That's the trick that makes this possible!
In order to cover the first quadrant r must go from 0 to \infty and θ from 0 to \frac{\pi}{2}. In polar coordinates, we have \frac{I^2}{4}= \int_{\theta=0}^{\frac{\pi}{2}}\int_{r=0}{\infty}e^{-r^2}rdrd\theta.

Since there is no θ in the integrand, we have immediately \frac{I^2}{4}= \frac{\pi}{2}\int_{r=0}^{\infty}re^{-r^2}dr.

Let u= r². Then du= 2rdr so rdr= du/2 (see why we needed that "r"?). Of course, when r=0,u= 0 and when t= \infty, so does u. In terms of the variable u, \frac{I^2}{4}= \frac{\pi}{4}\int_{u=0}^{\infty}e^{-u}du. That's easy to integrate and we have \frac{I^2}{4}= \frac{\pi}{4} so I^2= \frac{\pi} and I= \sqrt{\pi}.

I forgot the square root, as several people pointed out!

 

[danne89]

How about \int_{- \infty}^{+ \infty} x dx

 

[hypermorphism]

How about \int_{- \infty}^{+ \infty} x dx

You can use the basic definition of the improper integral over the reals as a limit to show that this expression diverges (represents no value in the reals).

 

[Galileo]

How about \int_{- \infty}^{+ \infty} x dx

That's an interesting case.

\lim_{\epsilon_{1,2} \to _{\pm}\infty}\int_{\epsilon_1}^{\epsilon_2}xdx=\frac{1}{2}\lim_{\epsilon_{1,2} \to _{\pm}\infty}\left(\epsilon_2^2-\epsilon_1^2\right)

The limit doesn't exist if \epsilon_1 and \epsilon_2 approach infinity in an arbitrary manner.

But if \epsilon_1=\epsilon_2=\epsilon:

\lim_{\epsilon \to \infty}\int_{-\epsilon}^{+\epsilon}xdx
does exist.
It's called the Cauchy principal value of the improper integral.

 

[kreil]

Just read all the posts, got stuck with detention for being late to school, lots of information to work on...

Galileo-

Integrals at infinity are defined by a limit, just like sequences or series.
Similarly, if the integral limit exists, the integral is called convergent.

\int_a^{\infty}f(x)dx=\lim_{t \to \infty} \int_a^t f(x)dx

If you could give me a simple example I would be able to fully grasp this.



Thank you halls of ivy for showing that to me, I will have to work on it a bit.

 

[Galileo]

Okay. A very simple case:

\int_{1}^{\infty}\frac{1}{x^2}dx=\lim_{t \to \infty} \int_1^t\frac{1}{x^2}dx=\lim_{t \to \infty}\left(-\frac{1}{x}\right]^t_1=\lim_{t \to \infty}\left(1-\frac{1}{t}\right)=1

Another one, using \frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}:

\int_{-\infty}^{+\infty}\frac{1}{1+x^2}dx=2 \int_{0}^{\infty}\frac{1}{1+x^2}dx=2\lim_{t \to \infty} (\arctan(t)-\arctan(0))=\pi

 

[t!m]

I think Gabriel's Horn ties in nicely in this discussion. For those who don't know, it is the shape generated for the revolution of 1/x around the x-axis for x>= 1. In doing the improper integrals, it has a finite volume of pi, but an infinite surface area. Blows my mind.

 

[dextercioby]

Good one...Congratulations! You managed to find the simplest possible...

Daniel.

P.S.That's a reply to post No.15.

 

[dextercioby]

That's an interesting case.

\lim_{\epsilon_{1,2} \to _{\pm}\infty}\int_{\epsilon_1}^{\epsilon_2}xdx=\frac{1}{2}\lim_{\epsilon_{1,2} \to _{\pm}\infty}\left(\epsilon_2^2-\epsilon_1^2\right)

The limit doesn't exist if \epsilon_1 and \epsilon_2 approach infinity in an arbitrary manner.

But if \epsilon_1=\epsilon_2=\epsilon:

\lim_{\epsilon \to \infty}\int_{-\epsilon}^{+\epsilon}xdx
does exist.
It's called the Cauchy principal value of the improper integral.

What about this one,Galileo:
\int_{-\infty}^{+\infty} x \ dx= 0

,as the integration of an odd function over an interval symmetric wrt the origin?


Daniel.

P.S.Are u sure that "a=-a" implies "a=0"??If you are,then u'll understand my arguments...

 

[Galileo]

What about this one,Galileo:
\int_{-\infty}^{+\infty} x \ dx= 0

,as the integration of an odd function over an interval symmetric wrt the origin?


Daniel.

P.S.Are u sure that "a=-a" implies "a=0"??If you are,then u'll understand my arguments...

Sorry, I don't see what you're hinting at.

And yes, ofcourse a=-a implies a=0.

BTW: One question mark will suffice.

 

[dextercioby]

What di u mean u don't see?

II gave a reason why that integral is zero:it's an integral from an odd function on a symmetric interval wrt the origin x=0.

Daniel.

 

[kreil]

\int_{-\infty}^{\infty}xdx=2\lim_{t\rightarrow{\infty}}\int_0^{t}xdx

=2\lim_{t\rightarrow{\infty}}1 ]_0^t=2(1-1)=0

figured I would take that chance to practice the formula galileo showed me

 

[dextercioby]

That's wrong,the function "x" is odd.U cannot split the interval into 2,because the function is not even...

Daniel.

 

[dextercioby]

And the primitive/antiderivative,is not one,but x^2/2.

Daniel.

 

[kreil]

bah

ok thanks

 

[Galileo]

What di u mean u don't see?

II gave a reason why that integral is zero:it's an integral from an odd function on a symmetric interval wrt the origin x=0.

Daniel.

Yes, which means

\int_{-a}^{+a} xdx=0
for all a\in \mathbb{R}.
It does not follow that:

\int \limits_{-\infty}^{+\infty} xdx=0

Since you are dealing with 2 limits:
\lim_{N_1 \to -\infty} \lim_{N_2 \to +\infty} \int \limits_{N_1}^{N_2} xdx=\frac{1}{2}\lim_{N_1 \to -\infty} \lim_{N_2 \to +\infty}\left(N_2^2-N_1^2\right)

This limit clearly doesn't exist. Therefore \int_{-\infty}^{+\infty} xdx is undefined.

What you are referring to is the following integral:

\lim_{N \to \infty} \int \limits_{-N}^{+N} xdx
which does equal 0. But this improper integral where the limits approach infinity in a nice symmetrical manner is called the Cauchy Principal value of the improper integral. So you see there is some subtlety involved.

 

[danne89]

Ahh. A little to advance for me. Isn't infinity a member of the reals?

 

[Muzza]

Nope, it's not.

 

[kreil]

The reals are composed of numbers, infinity is not a number but rather a concept