# Integration to infinite

1. Aug 26, 2010

### Mentallic

I want to understand intuitively why it seems to be that when evaluating an integral at infinite, it only seems to give a finite value for exponential functions that limit to zero as it approaches infinite. Others such as trigonometric and polynomial functions don't give finite values.

e.g.

$$\int_0^{\infty}e^{-x^2}dx=\frac{\sqrt{\pi}}{2}$$

$$\int_0^{\infty}\frac{dx}{x}=\lim_{k\to\infty}ln(k)$$

$$\int_0^{\pi/2}tan(x)dx=\int_0^{\infty}\left(\frac{\pi}{2}-tan^{-1}x\right)dx=\frac{1}{2}\lim_{k\to\infty}ln(k^2+1)$$

All these functions create the same shape that is, they tend to zero as x tends to infinite, but the difference is that the area under the graphs of the 2nd and 3rd between the x-axis and the function doesn't add to a finite value while it does for the first. How can I understand this intuitively?

2. Aug 26, 2010

### JJacquelin

No, the three functions given as an example doesn't create the same shape :
exp(-x²) tends to 1 as x tends to 0 and tends to 0 as x tends to infinity. So exp(-x²) never is infinite on the range of integration.
1/x tends to infinity as x tends to 0 and tends to 0 as x tends to infinity. So 1/x is infinite on the border x=0 of integration.
tan(x) tends to 0 as x tends to 0 and tends to infinity as x tends to pi/2. So tan(x) is infinite on the border x=pi/2 of integration.
It isn't always so simple. For example :
x/(x+1)^3 tends to 0 as x tends to 0 and tends to 0 as x tends to infinity. The integral from x=0 to x=infinity is convergent and = 1/2
x/(x+1)^2 tends to 0 as x tends to 0 and tends to 0 as x tends to infinity. The integral from x=0 to x=infinity isn't convergent (so, is infinite).

3. Aug 26, 2010

### Mentallic

Wait, I made a mistake. For the 2nd integral I intended to make it from 1 to infinite, since the function 1/x has a symmetry about the line y=x.

Oh yes those were the terms I was looking for. I want to know the criteria required for an integral to be convergent when its limit tends to 0 as it approaches infinite.

That's a nice example! It seems like it extends beyond the scope of only exponential functions being convergent. Actually at any point for a rational function where the degree of the denominator is more than 1 more the numerator, it will be convergent. This I don't understand intuitively so much either, it approaches zero at a faster rate of course, but what makes it so that x-1 diverges while x-1.01 converges.

4. Aug 26, 2010

### Vardd

I've asked myself this questions a few years ago, and It is not be the official answers but it helped me understand the "Why".

Integral(0 to infinity) ( 1/x) dx = Ln(x) +C , where C = 0.

So we can clearly see that as the n increase, it doesn't converge to any number.
An other way of seeing it, you can find the Series for 1/x. It is the Harmonic Series. Simply try to add the sum of 1/1 + 1/2 + 1/3 + 1/4 ..etc. Does it Converge?

Now, Let's try the samething but with x^(-1.01)

Integral( 0 to infinity) (x^(-1.01)) = -100/x^0.01 + C , where C = 0.
If we take the limit, we can already see the difference.

Hope it helps.

(Sorry for no LaTeX, I've tried to make it works, but after 10 min of unsuccessful attempts, i just wrote them down)

5. Aug 26, 2010

### Bohrok

Don't forget definite integrals such as $\int_0^1$ 1/√x dx which is also improper/infinite at the left limit yet convergent.

One definite integral I find fascinating is
$$\int_0^\infty\frac{1}{\sqrt{x}(x + 1)}dx = \pi$$
because it's improper in different ways at both limits of integration, yet it's convergent. I've often wondered if there are any other such integrals that also converge.

6. Aug 27, 2010

### JJacquelin

There are as many such integrales as you want. An infinity in fact. Some improper integrals converge, other doesn't converge. No wonder at all.
Like many topics en maths, the convergence of improper integrals or infinite series is a field of studies with a background of knowledge to learn.

7. Aug 27, 2010

### robert Ihnot

The matter of the second question can be seen in the sum: $$\sum_{i=1}^{\infty}\frac{1}{i}\rightarrow \infty$$ though the sum of the recipecals of squares converges to $$\frac{\pi^2}{6}$$

There are tests to determine convergence.

Last edited: Aug 27, 2010