use the subtitution rule
- Thread starter Jonkies
- Start date
- Joined
- Apr 12, 2005
- Messages
- 11,339
Well, you'll have to suggest what it is you did try and why you think it didn't work.
At x = 0, that argument in infinite. Will ANYTHING work? This is an important question.
At x = 1, that argument changes sign. Will that confuse anything?
At x = 0, that argument in infinite. Will ANYTHING work? This is an important question.
At x = 1, that argument changes sign. Will that confuse anything?
D
Deleted member 4993
Guest
Is your problem:
\(\displaystyle \int^{\infty}_0 ln\left[\frac{x}{1+x^2}\right]dx\)
.................................or
\(\displaystyle \int^{\infty}_0 \frac{ln(x)}{1+x^2}dx\)
Please clarify.
- Joined
- Apr 12, 2005
- Messages
- 11,339
Let's see...
[math]u = \dfrac{1}{x}[/math]
[math]du = -\dfrac{1}{x^{2}}\;dx[/math] <== Your version has the wrong sign and is missing some notation.
[math]\int^{\infty}_{0}\dfrac{-ln(u)}{u^{2} + 1}\;du[/math] <== It is not clear at all how you managed simply 1/x in your integrand. 1) Shouldn't that be something in terms of "u"? 2) Where did the logarithm go? 3) How did you lose the "1" in the denominator. Please be MUCH more careful with your algebra. If you get through all the algebra correctly, lo, and behold, we didn't get anywhere at all. That substitution resulted in essentially the same problem. That's no good.
Technical Point: If you know u > 0, there is no need to write |u|.
You have glibly written "ln(0)". That's no good. This goes back to my original questions. Why aren't you worried about convergence? "ln(0)" doesn't mean anything.
In your last two statements, it appears that [math]\infty[/math] has turned into "1"? How did that happen under the transformation 1/x? And how did zero remain zero under that same transformation? Or, was it [math]\infty[/math] turned into 0 and 0 turned into 1? I can't tell.
Technical Point: This is a definite integral. What it "+C" doing there at the end?
Why are we constrained to Substitution? Frankly, "By Parts" looks more compelling, but we still have that Convergence problem.
Why don't we try a simpler problem and see if we are getting the concept of Substitution? Maybe [math]\int\dfrac{2x}{x^{2}+1}\;dx[/math]?
[math]u = \dfrac{1}{x}[/math]
[math]du = -\dfrac{1}{x^{2}}\;dx[/math] <== Your version has the wrong sign and is missing some notation.
[math]\int^{\infty}_{0}\dfrac{-ln(u)}{u^{2} + 1}\;du[/math] <== It is not clear at all how you managed simply 1/x in your integrand. 1) Shouldn't that be something in terms of "u"? 2) Where did the logarithm go? 3) How did you lose the "1" in the denominator. Please be MUCH more careful with your algebra. If you get through all the algebra correctly, lo, and behold, we didn't get anywhere at all. That substitution resulted in essentially the same problem. That's no good.
Technical Point: If you know u > 0, there is no need to write |u|.
You have glibly written "ln(0)". That's no good. This goes back to my original questions. Why aren't you worried about convergence? "ln(0)" doesn't mean anything.
In your last two statements, it appears that [math]\infty[/math] has turned into "1"? How did that happen under the transformation 1/x? And how did zero remain zero under that same transformation? Or, was it [math]\infty[/math] turned into 0 and 0 turned into 1? I can't tell.
Technical Point: This is a definite integral. What it "+C" doing there at the end?
Why are we constrained to Substitution? Frankly, "By Parts" looks more compelling, but we still have that Convergence problem.
Why don't we try a simpler problem and see if we are getting the concept of Substitution? Maybe [math]\int\dfrac{2x}{x^{2}+1}\;dx[/math]?