help with this difficult improper Integral?

Question

my teacher said that the correct way of solving is too long (more the 2 complete pages)
be careful it's not as easy as you see at first it's a improper Integral

Infinity
∫ [(1-x) / x] dx
0

Answer 1

This integral does not actually exist. Proof: since the function is unbounded at zero and the other limit is infinity, this integral is improper at both ends. So it must be written as:

[a→0⁺, b→∞]lim [a, b]∫(1-x)/x dx

Now evaluating the proper integral:

[a→0⁺, b→∞]lim [a, b]∫1/x-1 dx
[a→0⁺, b→∞]lim ln |x| - x |[a, b]
[a→0⁺, b→∞]lim ln |b| - b - ln |a| + a

Now, [b→∞]lim ln |b| - b = -∞ (since b grows much faster than ln |b| and [a→0⁺]lim -ln |a|+a = ∞, so this limit is of the indeterminate form -∞+∞. This means that the limit can change depending on how fast a grows relative to b. For any real number r, we can force this to converge to r by letting a=be^(-b-r), so that it becomes [b→∞]lim ln |b| - b - ln |be^(-b-r)| + be^(-b-r) = [b→∞]lim ln |b| - b - ln |b| - ln |e^(-b-r)| + be^(-b-r) = [b→∞]lim -b-ln |e^(-b-r)| + be^(-b-r) = [b→∞]lim -b+b+r + be^(-b-r) = [b→∞]lim r + be^(-b-r) = r. Since there is no restriction on how fast a converges relative to b, however, this means that the limit truly does not exist, and neither does the improper integral.

Answer 2

there's slightly a "trick" which you would be able to hire right here. considering the fact that 2?(x) is non-damaging everywhere in this era, set u = 9ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c7. Then du = a million/[9ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c7] d9ba49493c14eb232109e2df225526c7, and 2?(x) = u². whilst 2?(x) = 0, u = 0 and whilst 2?(x) = a million, u = a million. With this variation of variables, the indispensable is a million . . . . . . . . . . .a million ? 6 ln9ba49493c14eb232109e2df225526c74u²2?(x) du = ? 19ba49493c14eb232109e2df225526c7 ln9ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c7u9ba49493c14eb232109e2df225526c7 du 0 . . . . . . . . . . 0 The final line follows from noting that 4u² = 9ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c7u9ba49493c14eb232109e2df225526c7², and the valuables of the log ln9ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c7u9ba49493c14eb232109e2df225526c7²2?(x) = 2?(x) ln9ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c7u9ba49493c14eb232109e2df225526c7. ok, now you combine by employing aspects. replace the decrease cut back of integration with a variable t 9ba49493c14eb232109e2df225526c70 < t < 19ba49493c14eb232109e2df225526c7, combine, then take the cut back as t->0+. ? 19ba49493c14eb232109e2df225526c7 ln9ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c7u9ba49493c14eb232109e2df225526c7 du = 19ba49493c14eb232109e2df225526c7u ln9ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c7u9ba49493c14eb232109e2df225526c7 - 19ba49493c14eb232109e2df225526c7 u. So a million ? 19ba49493c14eb232109e2df225526c7 ln9ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c7u9ba49493c14eb232109e2df225526c7 du = 19ba49493c14eb232109e2df225526c7ln9ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c7 - 19ba49493c14eb232109e2df225526c7 - 9ba49493c14eb232109e2df225526c719ba49493c14eb232109e2df225526c7t ln9ba49493c14eb232109e2df225526c79ba49493c14eb232109e2df225526c7t9ba49493c14eb232109e2df225526c7 - 19ba49493c14eb232109e2df225526c7t9ba49493c14eb232109e2df225526c7. t Take t to 0 from the appropriate, and you arrive at your answer.

Answer 3

The integral is the sum of a positive area (x = 0..1) and a negative area (x = 1..infty). In order to be able to compare them, I flip the positive area in the line y = x - 1, so that we get

∫1/x dx + ∫(1-x)/x dx
= ∫1/x dx - ∫(1 - 1/x) dx

both integrals running from 1 to infinity.

Now suppose x > 3. Then
... 1/x < 1/3
... 1 - 1/x > 2/3
So 1/x - (1 - 1/x) < -1/3.

Therefore, if we integrate from 3 to infinity,
∫1/x dx - ∫(1 - 1/x) dx < ∫-1/3 dx = -oo.

Since on the interval between 1 and 3 both integrals are finite, we conclude that the total integral is -oo.

Answer 4

Int (1-x)/x dx = lnIxI -x +C
Int =lim (lnIzI-z -lnIwI +w )as z=>infinity and w=>0
=lim ln Iz/wI-z
As z and w have no restrictions you can choose the way how one tends to 0 and the other to infinity.So the integral does not converge or could make to converge

Answer 5

The area enclosed between y=0 and y=(1-x)/x is infinite.