# Somebody slap me

1. May 9, 2005

### Hurkyl

Staff Emeritus
Nothing enlightening in this post, or even informative. Just the overwhelming need to complain about my own stupidity.

I've been staring at this integral all day:

$$\int_{-\infty}^{\infty} f(x)^{n-1} f'(x) dx$$

*sigh*

It doesn't dawn on me that this is a simple u-substitution until the drive home after a nice bridge session. Admittedly, f wasn't something with which I was eminently familiar -- it was $1 + \mathop{erf} (x / \sqrt{2})$, but still, there's no excuse!

But it gets worse. While I was working out how to set up the integral, I had already deduced the key step that would allow me to set up an easily solved recursive formula for my integral. I enjoy solving things with recursion. But did this dawn on me? Noooo... not until my drive home tonight.

*sigh*

But wait, my stupidity doesn't even end there! Before I was even interested in setting up and solving this integral, I had already figured out how the value behaves as n grows! (And the behavior as I adjust n was my primary interest)

But do I remember that? Nooo... not until the drive home today.

*sigh*

Sorry I took up so much space for this rant!

2. May 9, 2005

### Math Is Hard

Staff Emeritus
I guess even geniuses like Hurkyl have off days. Makes me feel a little better!

3. May 9, 2005

### Hurkyl

Staff Emeritus
The best part about it is this -- the constant out front of that integral had an n in it, and I actually thought to myself:

"Neat -- n (1 + erf (x/√2))n-1 looks like part of a derivative."

:grumpy:

4. May 10, 2005

### mathwonk

just to put things in perspewctive, heres a little one that stumped most of my class:

differentiate x^e (wrt x).

another puzzle for me was to decide whether to give half credit for

saying the derivative of x^(ln(x)) is ln(x)[x^(ln(x)-1)]. (work out the answer).

you sort of have to be a mind reader to know for sure.

5. May 10, 2005

### saltydog

So you have:

$$f(x)=[1+\frac{2}{\sqrt{\pi}}\int_{0}^{\frac{x}{\sqrt{2}}} e^{-u^2}du]$$

Then:

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

So that:

$$\int_{-\infty}^{\infty}[f(x)]^{n-1}f^{'}(x)dx=\int_{-\infty}^{\infty}[f(x)]^{n-1}e^{-\frac{x^2}{2}}dx$$

Ok, now what?

6. May 10, 2005

### krab

salty: The integral is $$\int^1_0 f^{n-1}df$$

7. May 10, 2005

### honestrosewater

Must I know what you're talking about if I can't slap you?

8. May 11, 2005

### saltydog

Would have taken longer than a drive home for me to see the integrand in that form. I don't see how the limits change to 0 and 1 but I'll spend some time on it.

Thanks.

9. May 11, 2005

### saltydog

No. It's not happening for me but that's all right. It's like that integral Daniel solved the other day. I wondered out of the 85 people who looked at it, who other than me couldn't solve it even after he gave the hint.

10. May 11, 2005

### saltydog

Alright, I let:

$$u=f(x)$$

$$du=f^{'}(x)dx=df$$

So, switching to u and changing limits I get:

$$f(\infty)=2$$

$$f(-\infty)=0$$

Thus I get:

$$\int_0^2 u^{n-1}du$$

When I integrate directly the integral (for n=5):

$$\int_{-\infty}^{\infty}[f(x)]^{n-1}f^{'}(x)dx$$

both numerically and analytically in Mathematica, I get 6.4. The above integral in u is also 6.4. Note the plot of the integrand. It's about 3 high by about 3 wide so say 9 or say somewhere between 5 and 10, I mean we're not building the shuttle or nothing. So I believe the upper limit of 2 is correct.

Edit: And you knew this from the get-go too Hurkyl, no driving around or nothing.

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Last edited: May 11, 2005