What do you get when you use integration by substitution where the subtitution is x = 1/t?
Hi everybody,
I have to integrate the function : ln(x)/(1+x²) between 0 and +inf.
I can't find the solution.
When I try an integration by part, I feel like running around in circles...
On the exercice statement, there is a suggestion: set x=(1/t) ...this don't help me...
The solution I have to find is 0.
Could you help me to resolve this?
Thanks in advance!
What do you get when you use integration by substitution where the subtitution is x = 1/t?
When I substitute:
x=1/t
=> dx=-1/t²dt
[0int+inf] [[ln(1/t)/[(1/t)²+1]]*-1/t²]*dt
= - [0int+inf] [ln(1/t)/(1+t)²]*dt
Then, I try a by party:
f= ln(1/t) -> df=-1/t
dg=1/(1+t²) -> g= atan(t)
=> ln(1/t)*atan(t)- - [0int+inf] (1/t)*atan(t)Then, i try a new by party on [0int+inf] (1/t)*atan(t) and I find a similar expression as the statement one...
(sorry for the "[0int+inf]" I don't know how to write it on a computer)
You did not change the limits of integration. We have x = 0 when t = ∞ and vice versa. So,
(swapping the limits and removing - before )
(since )
which is exactly the same integral as you started with, but with the opposite sign.
Click "Reply With Quote" to see the LaTeX code.
Also, here is a nice online French-English dictionary that I used to study French.
Thanks for the nice advices!
We find an exact opposite integral after x=1/t substitution, ok but does it prove immediatly as compared to the first integral that the result is 0?
Do I have to resolve the integral (that I can't manage) to show that?
Yes, it does, but you may need to verify some auxiliary facts. First, the statement that we use is
.
To apply this statement to the integral, you need to show that , or that the integral converges, as John said in post #3. Also, you need to double-check that integrating by substitution works for improper integrals (I just don't remember the relevant theorems) and that it is done correctly (for example, x = 1/t = 0 corresponds to t = +∞ and not t = -∞). In general, the level of strictness depends on the course you are taking.
Note also that instead of proving , post #3 suggests showing that
using the same substitution.