Approximation methods that can be applied

[hedlund]

How do I find integrals like
\int_{a}^{b} \left( x^2 + 1 \right)^2 \ dx. This one is easy, since I can just turn it into \int_{a}^{b} \left( x^4 + 2x^2 + 1 \right) \ dx. But what if it would say \int_{a}^{b} \left( x^2 + 1 \right)^{40} \ dx? What technique should I use?

 

[devious_]

There are certain approximation methods that can be applied.

 

[matt grime]

Or substituion methods to obtain a (possibly hyperbolic) trig integral. Often you'll end up with trying to integrate something that we can write as:

I(n) = int f(x)dx

where f( x) is some function of x, and the n indicates some exponent or coefficient (the 40 in your case), then doing something like integration by parts you'll end up with

I(n-1) + something= I(n)

and you get an iteration going that let's you figure out how to solve the integral.

 

[matt grime]

I think you need to fix that latex.

 

[Muzza]

Yeah, I tried to (and then I deleted my post before I noticed that you had replied to it. Doh). LaTeX doesn't seem to like the combination of an integral sign, sum sign and a binomial coefficient... Let's try again:

\int (x^2 + 1)^{40} dx = \int \sum^{40}_{k=0} C^{40}_{k} x^{2k} dx = \left( \sum^{40}_{k=0} \frac{C^{40}_{k}}{2k + 1} x^{2k + 1} \right) + D.

Actually calculating those binomial coefficients is going to suck though :P

 

[Theelectricchild]

Leaving it as a series solution is absolutely fine--- consider

\int e^{x^2} dx

There is no substitution you can use, but you can use the general series form of e^x and plug in x^2 and then integrate the sum as a whole, and leave you answer as that.

 

[JonF]

this maybe silly, but wouldn't a trig sub then a integral table work for something like that?

 

[Dr Transport]

try the trig substitution for the original problem, let
x = \tan(y), then the portion in the parentheses becomes \sec(y)^2 from there you can make the substitutions and complete the integral. As for the e^{x^{2}} over a finite interval, you have to use an integral table because it can only be done numerically, if the limits are infinite in extent, then the integral is \frac{\pi}{2},

 

[Theelectricchild]

\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}

 

[HallsofIvy]

Yes, but the theelectricchild's point was that there is no "elementary" anti-derivative for \[e^{x^2}\]. The sad fact is that the great majority (in a very specific sense, "almost all") of nice "elementary" functions, even though they have an anti-derivative, the anti-derivative is not an "elementary" function.

 

[Dr Transport]

I forgot the square root...

 

[Theelectricchild]

Hehe its ok, that's one of the first places I have seen a "squareroot" of pi... its an integral you can actually solve using a double integration step.

\int_{0}^{\infty} e^{-x^2} dx

Convert to DI problem:

{(\int_{0}^{\infty} e^{-x^2} dx)}^2

=(\int_{0}^{\infty} e^{-x^2} dx) (\int_{0}^{\infty} e^{-y^2} dy)

=\int_{0}^{\infty}\int_{0}^{\infty} e^{-x^2}e^{-y^2} dxdy

=\int_{0}^{\infty}\int_{0}^{\infty} e^{-{(x^2+y^2)}} dxdy

From here using a Change of Variables step or noticing the region we are integrating is simply a semicircular arc, we can use polar coordinates:

=\int_{0}^{\frac{\pi}{2}}\int_{0}^{R} e^{-r^2}r drd\theta

Where we take the limit of this as R --> infinity and the extra r in the integrand is from the Jacobian when switching to polar coordinates (proven in the change of variables theorem)

And NOW we can use a U-substitution to easily solve this iterated double integral:

{(\int_{0}^{\infty} e^{-x^2} dx)}^2 = \frac{\pi}{4}

So \int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}

It seems like we can solve quite a few "definite" or "improper" integrals of non elementary functions using multivar. calculus. Does anyone know any other methods? Thanks.

 

[Theelectricchild]

I like integrating things.

 

[Divergent13]

Where did the R come from again?

 

[Theelectricchild]

Well taking an improper integral requires you to take the limit of the evaluated integral as some letter in place of infinity goes to it--- I didnt learn to do limits in TeX yet Ill show it later.