I need to evaluate the following expression for a bonus assignment:

Eavg = 0.453*int[E*exp(-1.036*E)*sinh(sqrt(2.29*E))*dE, 0..infinity]

(See attached image)

I need to do this using only first principles (no integration tables, no math programs, etc) and the relation given after the arrow at the top of the pic. I'm learning how terrible I am at complex integration, been a while since I've had to do it by hand and lewisje helped me out

You can see my thought process here but everything past where I divide it into two integrals is likely wrong. I didn't even bother to include the two more pages I have written out, all wrong anyway

Any help is greatly appreciated, and if you can point me in the right direction I'll buy you a beer or three.

10/18/2011 3:11:57 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.

[Edited on October 19, 2011 at 12:04 AM. Reason : hey at least I simplified it

10/19/2011 12:04:18 AM

You, sir are one hell of a gentleman. That all makes sense upon first inspection but im gonna finish up the rest of the required parts of the assignment before I tackle this beast. Will report back. Infinite thanks for your help...I'm prepared to fulfill that beer promise should you desire

10/19/2011 1:42:33 AM

I think by substitution you can solve this. Tomorrow I'll take a bit harder look.

And I know I have solved this integral before. But I'm getting dumb in my old age but I do know the average energy of a fission neutron is about ~2 MeV.

10/19/2011 1:46:23 AM

yeah, it's somewhere just shy of 2MeV

which is FINITE of course

alas, my current solution is not. I suppose it has something to do with not evaluating the definite integrals after each IBP operation, but looking back at it, it blows up at the very beginning anyway.

but i feel much better about it thanks to lewisje's genius

here's my last attempt:




someone mentioned putting the integral into an exp(ax^2+bx+c)dx using the hyperbolic identity and then using the error function.

While that may work, the bonus is due in an hour and there's no way I can chug through all of that. Hey, at least I'm confident in the exam itself.

Thanks again for all the help, even if I don't get much credit for the bonus I still learned a couple things

10/19/2011 10:06:35 AM

well if the error function is okay (as opposed to elementary functions only) then there's a much easier integration by parts to be done

10/19/2011 7:39:28 PM

i'm assuming the error function is not okay, not sure why he thought it was

10/19/2011 7:52:32 PM

maybe it's a casus irreducibilis, where you have to use non-elementary functions even though the final answer is expressible with only elementary functions of either algebraic or well-known transcendental numbers (like e and pi)

10/19/2011 10:16:14 PM

the answer is 1.9836

I know you didn't want that, but I couldn't help myself. I think my answer would be late by this time anyway and I'm sure whatever you needed it for has passed by now.

But srsly, Maple 10 that I have wouldn't crack it symbolically. Wolfram Alpha, on the other hand, was probably jealous and gave a real indefinite integral.

http://www.wolframalpha.com/input/?i=int+x*exp%28-1.036*x%29*sinh%28sqrt%282.29*x%29%29+dx

For real though, this isn't something to do by hand. You need to have a FIRM grasp on the erf() function, and I don't think anyone even mentioned that in this thread.

Just being honest, you were all probably wasting your time on this problem.

[Edited on November 10, 2011 at 4:57 PM. Reason : ]

11/10/2011 4:57:15 PM

Actually, the solution he gave didn't involve an error function. Apparently after a few key substitutions you had to complete the square and make some new cognitive leap followed by about 10 more sub-integrals, each with their own trickiness. Apparently only 2 kids in our class got it, and I remember them working on it for about 3 weeks straight. Me trying to bang it out in the day before it was due was a little optimistic but unfortunately I just didn't have that much time to devote to a bonus problem on an exam I was confident about. Made an A on the exam but it still aggravates the hell out of me that I didn't finish the bonus.

[Edited on November 10, 2011 at 6:42 PM. Reason : appreciate all the input though, I'm serial]

11/10/2011 6:39:36 PM

Quote :

"Actually, the solution he gave didn't involve an error function. "

Ah darn it, I guess it can go out when doing the definite integral. there's certainly no way to get anything less complicated than what I had unless you do integration by parts using the definite integral method.

answer:

alpha = 2.29
beta = 1.036

Eavg = 0.453 * 1/8 * sqrt(Pi*alpha) * (6 beta + alpha) * exp(alpha / (4 beta) ) / beta^(7/2)

= 1.98363

11/10/2011 7:35:19 PM

Yeah. I've noticed a trend with NE professors here...they love to give you basic principles, an answer, and tell you to do everything in between.

fine by me, but sometimes it's an obnoxious amount of work

11/10/2011 8:12:08 PM

Quote :

"Apparently after a few key substitutions you had to complete the square and make some new cognitive leap followed by about 10 more sub-integrals, each with their own trickiness."

I got to the completing the square part so at least I was on the right track. Of course in the real world I would have just written a numerical routine in ten minutes and be done with it.

11/14/2011 8:47:07 PM

what class is this for?

11/15/2011 11:31:22 AM

Quote :

"Of course in the real world I would have just written a numerical routine in ten minutes and be done with it"

^^ I realize this is old, but that's not necessarily true for a couple of reasons. Run-time can be a huge issue, especially in real-time software. So, any time you can avoid numerical integration is good. Also, often times people want to see analytic result because it can provide much greater insight into the problem. With an analytic expression you can derive analytic derivatives, which will greatly increase performance in optimization routines.

We actually do a lot of analytic integration and computations to increase performance for our simulations. If we can't get it analytic, then we have to do an offline numerical scheme and build a look-up table.

[Edited on February 6, 2012 at 9:28 PM. Reason : .]

2/6/2012 9:27:02 PM

^^ was for NE301 with AK

now in Anistratov's class and everything prior seems like so much child's play

[Edited on February 7, 2012 at 2:32 PM. Reason : ^ interesting insight ITT]

2/7/2012 2:31:52 PM

damn, I don't remember crap like that from 301 with Hankins.

2/7/2012 2:33:48 PM

b = 0

whats the big hoopla?

I'm Krallum and I approved this message.

2/7/2012 10:12:03 PM