Integration by parts constants

[Yegor]

It's not homework, but i think it can make someone think a little.
$$\int\frac{dx}{x}$$
Take it by parts.
If you'll be as careless as me you can make a discovery :rofl:

 

[arildno]

Set v'=1/x, u=1.

 

[HallsofIvy]

I'm clearly not understanding this (and it really belongs in the Calculus section rather than Homework).

Using arildno's suggestion: $$dv= \frac{1}{x}dx$$, u= 1 just means that you have to integrate $$\int\frac{dx}{x}$$ directly.

The only other possibility is dv= dx, $$u= \frac{1}{x}$$ so that v= x, $$du= \frac{-1}{x^2}$$. Then integration by parts gives $$\int\frac{dx}{x}= (x)\frac{1}{x}- \int\frac{-xdx}{x^2}= 1+ \int\frac{dx}{x}$$ which is certainly true allowing for different constants of integration.

 

[arildno]

It was a trick question, HallsofIvy:
He wanted to see if someone besides himself made a sign error so that you "get", say:
$$\int\frac{dx}{x}=1-\int\frac{dx}{x}\to\int\frac{dx}{x}=\frac{1}{2}$$

 

[dextercioby]

Nope,my guess is it must have been 1=0.

Daniel.

 

[Jameson]

This can applied to any basic integral. Example:

$$\int xdx$$

I'll take this by parts.

dv = x
u = 1

$$\int udv = uv-\int vdu$$

$$\int xdx = 1*\frac{x^2}{2}-\int \frac{x^2}{2}*0dx + C_1$$

The $C_1$ from the first integral and the $C_2$ from $\int 0 dx$ can just be called $C$

So the resulting answer is the same:

$$\int xdx = \frac{x^2}{2} + C$$

I'm sorry to say I find this a little pointless...

 

[arildno]

Of course you can do that, Jameson.
I wasn't particularly serious about the whole thing, since the original post was, after all, rather silly.

 

[Jameson]

I wasn't directing that at you... it was to the OP. I think it's agreed that this is silly.

 

[arildno]

Allright.
But, to take the thread into a bit of defense:
I'm sure we can recognize Yegor's feeling:
We do a calculation over and over again and get some complete nonsense out of it.

Then, slapping our head, we realize what an idiotic mistake we've made, and will start laughing over the whole matter .
I don't think Yegor ever meant it to be serious, since he had a :rofl: in his post, and perhaps want to share the joke with others.

 

[Yegor]

Of course i disappoint you, but i meant exactly what Daniel wrote (0=1).
I just had some associations with $$\int e^{ax}\sin(bx)dx$$, where we have to make similar trick. And that's why posted it here.
Don't be so strict, please, if someone isn't so brainy as you and pays attention to silly things.

 

[dextercioby]

This is for Arildno: :tongue: My intuition was simply sublime

Daniel.

P.S.Yegor,only smart people learn from mistakes.Welcome aboard !

 

[arildno]

Yes, you were absolutely divine today, daniel.

 

[arildno]

Of course i disappoint you, but i meant exactly what Daniel wrote (0=1).
I just had some associations with $$\int e^{ax}\sin(bx)dx$$, where we have to make similar trick. And that's why posted it here.
Don't be so strict, please, if someone isn't so brainy as you and pays attention to silly things.

Okay, sorry that I phrased myself in a manner which seemed contemptuous towards you. That was not my intent at all, but evidently the result anyway.

Sorry about that.