Guineafowl's latest activity

Now, at my stage, the way I know to solve a definite integral is do exactly this. In the real world, are you saying this isn’t always...

Stripping away the ##\lambda##, researching examples of the type: $$\int_0^\infty u^n e^{-u} du$$ I found that this is the gamma...

Right, I tried to knock your question into bits, and accidentally found something called the Gamma function. First, the origin of the...

I see. I assumed the wrong sign must have crept in during the difficult integration bit, not the easy final bit. Putting my water wings...

All three terms have minus signs there, when you expand it out.

Strewth, are we back to basics again? You may remember me from such threads as...

What you posted in post #7 is consistent with the integral calculator answer. It looks like you got a minus sign confused in the final...

The integral-calculator.com has it like this: $$-\frac{\left(2x^{2} + 2ax + a^{2}\right) \mathrm{e}^{-\frac{2x}{a}}}{4a}$$ Versus my...

Combining the result in post #4, which for some reason isn’t displaying, with Equation 1 of post #3: $$\frac{-ax^2 e^{-2x/a}}{2} -...

Beyond my pay grade, I think ;)

I'm not sure why you are calculating an indefinite integral. The first rule of mathematical physics is that the first term in a parts...

Tackling: $$\int -ae^{-2x/a} x dx$$ $$-a \int xe^{-2x/a}dx$$ IBP again: ##u=x, \frac {du}{dx}=1, \frac {dv}{dx}=e^{-2x/a}, v=\frac...

Ok, how about simplifying to: $$\int x^2e^{-2x/a}dx$$ Using the integration by parts as above, ie ##u=x^2, \frac {du}{dx}=2x, \frac...

Solving the integral for a wave function: $$1 = |C|^2 \int_0^\infty \frac{x^2}{a^2} e^{-2x/a} dx$$ Shunting out the ##\frac{1}{a^2}##...

I don’t know, but some searching suggests Zigbee latency is a bug, not a feature. Troubleshooting involves the usual suspects, eg...