Splitting an integral range to handle singularities

[GarethB]

I am required to write a program that uses Simpson's rule to evaluate ∫t**-2/3(1-t)**-1/3 dt from limits t=0 to t=1. The questions gives a hint to split the integral into two parts and use a change of variable to handle the singularities.
I really don't know where to begin. Is the choice of the value of t of where to split it arbitrary?
How would you handle the change of variables?

 

[gulfcoastfella]

It's a bit difficult to interpret your function the way you've entered it. Could you re-enter it using Latex Reference (click the $\Sigma$ symbol above the comment box you're typing in for Latex Reference formatting). Thanks.

 

[sunjin09]

since the singularities are the endpoints, what you can do is convert the domain of integration to [0, pi] and use a transformation, t=cosτ, so that dt=-sinτ dτ and hopefully the sinτ can cancel the singularity at the endpoints.

 

[GarethB]

Ok let me try to re-enter this;
the integral I am trying to compute is;
[0]\int[/1][t]^{-2}[/3][(1-t)]^{-1}[/3]dt
I have never used latex before so bare with me if this is a disaster

 

[GarethB]

ok sorry that clearly didn't work!

 

[Mark44]

I am required to write a program that uses Simpson's rule to evaluate ∫t**-2/3(1-t)**-1/3 dt from limits t=0 to t=1.

Other people responding to your post are confused about what the integrand is. If you write it inline, you NEED parentheses around your fractions.

As best as I can tell, this is your integrand:
t^(-2/3) * (1 - t)^(-1/3)

In a little nicer form (and not using LaTeX), this would be
t^-2/3(1 - t)^-1/3

I'm using the HTML tags for exponents here. They are available when you click Go Advanced just below the text input pane. For exponents, click the X² icon.

An even nicer form for your integral uses LaTeX.
$$\int_0^1 t^{-2/3} ~ (1 - t)^{-1/3}dt$$

If you click the integral expression, you can see the LaTeX script I used to get this integral.

 

[GarethB]

Yes that's what I have been trying to write all along; thanks for clearing that up now we can proceed with trying to solve this! Clearly there are singularities at either end. How could you split range of the integral into two parts and then use a different variable change in each?

 

[GarethB]

Every variable change I have tried just swaps where the singularities are.

 

[GarethB]

I have tried x=1/t as well as x=sin(t). The singularities remain.

 

[sunjin09]

ok sorry that clearly didn't work!

I assume you are referring to my suggested change of variable t=cosτ. Well as I tested it worked for the t=1 singularity and not for the t=0 one. What you can do is split the integral into (0,1/2) and (1/2,1), my trick works for the latter. Similar trick can be applied to the first interval, although I admit there must be a nicer way that handles both ends at the same time.

 

[GarethB]

No I was referring to my attempt at using latex (that clearly didn't work) I'm sorry I hope I didn't offend you. Anyway back to the problem.
I tried your suggestion of using t=cos(x). Surely then the upper limit will be zero, (since cos(0)=1) in which case the term (1-cos(x))^-1/3 will surely be undefined at the upper limit? Or am I missing something? Would you mind showing me how you solved it for the upper bound?

 

[gulfcoastfella]

Ignore this post, I was in the wrong thread...

 

[sunjin09]

No I was referring to my attempt at using latex (that clearly didn't work) I'm sorry I hope I didn't offend you. Anyway back to the problem.
I tried your suggestion of using t=cos(x). Surely then the upper limit will be zero, (since cos(0)=1) in which case the term (1-cos(x))^-1/3 will surely be undefined at the upper limit? Or am I missing something? Would you mind showing me how you solved it for the upper bound?

The (1-cos(x))^-1/3 blows up as x→0, but dt=-sin(x)dx, the sin(x) factor CANCELS the singularity, i.e., (1-cos(x))^-1/3*sin(x) is well behaved as x→0, the resultant integral can be evaluated in $x\in(0,\pi/3)$, corresponding to the t integral over (1/2,1), more specifically
$$\int_{1/2}^1 t^{-2/3}(1-t)^{-1/3}dt=\int_{\pi/3}^0 (\cos x)^{-2/3}(1-\cos x)^{-1/3}(-\sin x)dx$$
$$=\int_0^{\pi/3}(\cos x)^{-2/3}(2(\sin\frac{x}{2})^2)^{-1/3}2\sin\frac{x}{2}\cos\frac{x}{2}dx =\int_0^{\pi/3}(\cos x)^{-2/3}2^{2/3}(\sin\frac{x}{2})^{1/3}\cos\frac{x}{2}dx$$
Similar tricks can be applied for the t integral over (0, 1/2) interval.