Different results in the apparently same improper integral (wolfram)

Question

I thought the both limits above would give the same results, but it doesn't. Can someone explain why?

Answer 1

You have by the change of variable $u=x^2$ $$\int_0^n \sin(x^2)^2\cos(x^2)^2dx=\frac{1}{16}\int_0^{n^{2}}\frac{1-\cos(4u)}{\sqrt{u}}du=\frac{n}{8}-\frac{1}{16}\int_0^{n^{2}}\frac{\cos(4u)}{\sqrt{u}}du $$
Now this show (as the integral $\int_0^{+\infty}\frac{\cos(4u)}{\sqrt{u}}du $ is convergent) that the integral $\int_0^n \sin(x^2)^2\cos(x^2)^2dx$ is divergent. Hence the first computation is not defined.

For the second, it is now clear by the formula above, again as the integral $\int_0^{+\infty}\frac{\cos(4u)}{\sqrt{u}}du $ is convergent.

Answer 2

These two expressions are different. To help you see why, we will examine the first expression. $$\lim_{n \to \infty}\frac{\int_0^{\infty}\sin^2(x^2)\cos^2(x^2)\;dx}{n}$$ As you probably know, the integral $\int_0^{\infty}\sin^2(x^2)\cos^2(x^2)\;dx$ is just $\lim_{t\to \infty}\int_0^{t}\sin^2(x^2)\cos^2(x^2)\;dx$. Hence, the first expression is just $$\lim_{n \to \infty}\frac{\lim_{t\to \infty}\int_0^{t}\sin^2(x^2)\cos^2(x^2)\;dx}{n}.$$ Now observe the second expression $$\lim_{n \to \infty}\frac{\int_0^{n}\sin^2(x^2)\cos^2(x^2)\;dx}{n}.$$ While these may look remarkably similar, they are not the same thing. This is because in the first expression, $t$ is independent of $n$. Hence, when $t$ tends to infinity, $n$ does not. However, in the second integral, both the upper limit of the integral and the denominator tend to infinity simultaneously.

This might be a bit difficult to grasp at first, but think about it this way. Assume the integral $\int_0^{\infty}\sin^2(x^2)\cos^2(x^2)\;dx$ is convergent. (I don't think that it is, however, for now, we will pretend that it is). Hence, when you take the limit as $n$ tends to infinity, you are really taking the limit of some constant over $n$ as $n$ tends to infinity, or if you prefer equations, you are taking the limit $\lim_{n \to \infty}\frac{k}{n}$, for sime constant value $k$. This is clearly $0$. Now, assume that$\int_0^{\infty}\sin^2(x^2)\cos^2(x^2)\;dx$ is divergent. If this is the case, the above limit is undefined, since the expression that we are taking the limit of was undefined in the first place. Note however, that still the numerator does not depend on $n$, it is just always undefined.

Now think about the second expression, indeed you are taking the limit of something over $n$ as $n$ approaches infinity, but this something does not remain constant as $n$ increases. To see this, simply think about what you will get when you evaluate the top integral, you will get some function in $n$. (If you don't see this, try evaluating the integral yourself.) This will change as $n$ increases, causing the integral to increase.

An elementary example of such an effect would be the limits $\lim_{n \to \infty} \frac{n^2}{n^2}$ and $\lim_{n \to \infty} \frac{lim_{k\to \infty}k^2}{n^2}$, the first integral is clearly $1$, since as $n^2$ in the denominator increases, the $n^2$ in the numerator increases accordingly, keeping the expression at $1$, while the second expression would be undefined since $k$ increases independently of $n$ and gives us an undefined expression.

I have not checked either expression to see if wolfram alpha evaluated them correctly. However, I hope you now understand that the reason wlofram alpha output two different results was because the two expressions that you put in are not equivalent. That is far more important in my opinion than whether or not wolfram alpha is correct.

I hope this helped.