Improper intergral, infinite bounds of integration

[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

 

[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.

 

[nick.martinez]

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.

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.

 

[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.

 

[nick.martinez]

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.

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.

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 because if so the question on the quiz didn't exlicitly ask for cauchys principal.

 

[nick.martinez]

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.

Also, when you say the function is asymmetric 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.

 

[SteamKing]

∞
∫ 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

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

 

[MathematicalPhysicist]

Also, when you say the function is asymmetric 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.

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.