Anyone know any clinical trials currently looking for people? I know AAI Pharma did a bunch of stuff 5-6 years ago, but can't seem to find anything for them now.

12/11/2011 6:31:12 PM

might be a thrifty Christmas in the bbehe household

12/11/2011 6:39:13 PM

Haha, nah, it was good money though. I used to do AAI before I left the area. It was awesome go there for 2 weekends, make 900-1500.

12/11/2011 6:43:22 PM

well, where do you live now?

12/11/2011 6:44:21 PM

Raleigh.

12/11/2011 6:44:54 PM

so this "leaving the area" is false... can we trust anything you say Mr. bay-bayhe?

12/11/2011 6:53:17 PM

please don't get him started on Italy JB

12/11/2011 7:42:29 PM

If you don't need teh money, then why the fuck would you want to do this?

12/11/2011 7:44:33 PM

because he's cheap as fuck

12/11/2011 7:45:49 PM

I had a roomie with nasty allergies who did a clinical trial for allergy meds...he apparently got the placebo and was fucking miserable the entire time.

12/11/2011 7:47:17 PM

Before my long spiel, I should mention that the part on your work after you started using u' and du' is wrong: The antiderivative of e^f(E), where f(E) is some arbitrary function of E, is not generally (1/f'(E))e^f(E), even though it is true that the antiderivative of f'(E)e^f(E) is e^f(E).



First consider the transformation E=x^2/2.29; then the limits remain the same, 1.036E=(518/1145)x^2, sinh(sqrt(2.29E))=sinh(x), and 0.453E dE=(453/1145)x^2 dx.

Then the integral becomes the integral from 0 to +infinity of (453/1145)x^2*e^(-(518/1145)x^2)*sinh(x) dx.

--------Begin Failed Solution Attempt--------
Then in integration by parts, let u=sinh(x) and dv=(453/1145)x^2*e^(-(518/1145)x^2) dx.

Then du=cosh(x), and to find v, do another integration by parts, letting U=x and dV=(453/1145)x*e^(-(518/1145)x^2) dx.

Then dU=dx and V=-(453/1036)e^(-(518/1145)x^2).

Therefore, v=-(453/1036)x*e^(-(518/1145)x^2)+(453/1036)Int(e^(-(518/1145)t^2) dt,0,x).

Then our original integral is lim(uv,x->+inf)-(uv)(0)-Int(v du,0,+inf), so keeping in mind that u=(1/2)e^x-(1/2)e^-x...
...hmm, I see a problem, it looks like one of the terms will go off to infinity!

Notice that one term will be sinh(x)*(453/1036)Int(e^(-(518/1145)t^2) dt,0,x), in the limit as x->+inf; by the relationship at the top, the right factor becomes a finite positive number, but the left factor becomes infinite.

This must mean that I have a problem with my method, because as you can imagine, sinh(k*sqrt(E)) for constant k grows asymptotically as quickly as e^sqrt(E), which is more slowly than e^(-1.036E) reaches 0; then their product reaches 0 as quickly as e^-E, which is faster than E grows, in fact so quickly that the whole integrand also reaches 0 in exponential time, so the integral should be finite.
--------End Failed Solution Attempt--------

Let's go back to the step where u=sinh(x) and dv=(453/1145)x^2*e^(-(518/1145)x^2) dx; instead let's try u=x*sinh(x) and dv=(453/1145)x*e^(-(518/1145)x^2) dx.
Then du=sinh(x)+x*cosh(x) dx and v=-(453/1036)e^(-(518/1145)x^2).
Then the antiderivative is -(453/1036)x*sinh(x)e^(-(518/1145)x^2)+(453/1036)int(sinh(t)e^(-(518/1145)t^2)+t*cosh(t)e^(-(518/1145)t^2),0,x).

Now for the first term in that "int"-egral...letting U=e^(-(518/1145)t^2) and dV=sinh(t) dt, dU=-(1036/1145)t*e^(-(518/1145)t^2) dt and V=cosh(t), so the antiderivative becomes cosh(x)e^(-(518/1145)x^2)+(1036/1145)int(t*cosh(t)e^(-(518/1145)t^2) dt,0,x).
Now for the second term in that same "int"-egral...letting w=cosh(t) and dz=t*e^(-(518/1145)t^2) dt, dw=sinh(t) dt and z=-(1145/1036)e^(-(518/1145)t^2), so the antiderivative becomes -(1145/1036)cosh(x)e^(-(518/1145)x^2)+(1145/1036)int(sinh(t)e^(-(518/1145)t^2) dt,0,x).

If we denote 1036/1145 by k, then the second sentence means "so the antiderivative becomes -(1/k)cosh(x)e^(-(k/2)t^2)+(1/k)int(sinh(t)e^(-(k/2)t^2) dt,0,x)" and the first sentence means "so the antiderivative becomes cosh(x)e^(-(k/2)x^2)+k*int(t*cosh(t)e^(-(k/2)t^2) dt,0,x)" (to make it all cleaner).
Now there is one more integration by parts to do; if W=cosh(t) and dZ=t*e^(-(k/2)t^2) dt, then dW=sinh(t) dt and Z=-(1/k)e^(-(k/2)t^2), so that integral becomes -(1/k)cosh(x)e^(-(k/2)x^2)+(1/k)int(sinh(t)e^(-(k/2)t^2) dt,0,x).
Substituting back, some terms cancel out in "the first sentence" to yield just the innermost integral, int(sinh(t)e^(-(k/2)t^2) dt,0,x); then adding that to "the second sentence" yields the entire expression in that "int"-egral that I referred to earlier: -(1/k)cosh(x)e^(-(k/2)t^2)+(1+1/k)int(sinh(t)e^(-(k/2)t^2) dt,0,x).

Now wrapping up all the integrations by parts and back-substitutions so far, the antiderivative is
-(453/1036)x*sinh(x)e^(-(k/2)x^2)-(1/k)cosh(x)e^(-(k/2)t^2)+(1+1/k)int(sinh(t)e^(-(k/2)t^2) dt,0,x); this last unresolved integral does not appear amenable to integration by parts, and I don't quite know how to attack it, but when you evaluate the limit as x->+inf, that becomes finite. As for the two terms outside the integral, a similar asymptotic analysis shows that they go to 0 as x approaches 0 or +inf, so really what you're left with is (1+1/k)int(sinh(x)e^(-(k/2)x^2) dx,0,+inf), whatever that is.

12/11/2011 7:47:34 PM

^
you should stop posting that....my head hurts

12/11/2011 10:16:37 PM

^^ I think I just snowcrashed...

12/11/2011 10:18:39 PM

^^^ You're a pretty horrible poster.

12/11/2011 10:32:55 PM