# Improper intergral, infinite bounds of integration

1. Mar 31, 2013

### nick.martinez

∫ x^3 dx =∫ x^3 dx+∫x^3dx
-∞

I split up integral and got (x^4)/4 and infinity when evaluating using limits. does the integral converge to zero or diverge to infinity minus infinity?

would really appreciate help answering this question.
Thanks

2. Mar 31, 2013

### MathematicalPhysicist

Well, because it's an antisymmetric function, from the graph you can see that the area from x>0 has the same magnitude as the area from x<0 but with a minus sign. Thus the integrals should cancel each other, but because their area is infinite this integral is ill defined.

3. Mar 31, 2013

### nick.martinez

So when do the areas cancel? Because on my last quiz i had this question.

∫[(x)/((x^2)+1)]dx
-∞

the antiderivative is:
(Ln|x^2+1|)/2 and i (Ln|∞+1|)/2-(Ln|∞+1|)/2

which gives ∞-∞.
by the way i got this answer by using limits. Some people in class were trying to say the function converges to zero but I disagree. Please elaborate if i am correct in this case on my quiz. and also when can i say ∞-∞ is equal to zero.

4. Mar 31, 2013

### MathematicalPhysicist

These sorts of itnegrals are ill defined, and don't converge not even to infinity.

Take for example:

$$\lim_{T\rightarrow \infty} \int_{-T}^{2T+1} x/(x^2+1) dx = \lim 1/2 ln(\frac{((2T+1)/T)^2 +1/T^2}{1+1/T^2})=ln 2$$
and you can change the value of the integral as you see fit, perhaps 3T+1 instead of 2T+1, etc.

You see you get different values for this integral.

5. Mar 31, 2013

### nick.martinez

not quite sure what is going on in the notation. why is the positve infinity not 2t+1 and the negative bound on -T. And is this cauchy's principal becuase if so the question on the quiz didn't exlicitly ask for cauchys principal.

6. Mar 31, 2013

### nick.martinez

Also, when you say the function is assymetric and so the areas should cancel, but since the limits of integration are infinite the function diverges. But when ∫x^3 dx we are essentially finding the area under the curve of (x^4)/4 which is an even function so the areas should not cancel because the function is symmetrical on both sides.

7. Mar 31, 2013

### SteamKing

Staff Emeritus
Infinity minus infinity is not equal to zero. You have a divergent integral.

8. Mar 31, 2013

### MathematicalPhysicist

When you have an expression of infinity-infinity you get what is called indeterminate value, as in it could be any value you wish.

In the integral of x^3, you can pick any parameter T which approaches infinity.

For example we may take the integral between -T and T and get in the limit that it equals zero, or we may take from -T and T+1 and get that it converges to infinity.