Recent content by Guineafowl

I’ve also seen it written: $$\int fg’ \,dx = fg - \int f’g \,dx$$ Maybe you could preface with ‘f and g are functions of x’ and save all the (x)s.

Thanks, that all makes sense. All I will say for the ##uv## version is, it’s briefer, and for the integral bits, there’s a pleasing symmetry to: $$\int u\frac{dv}{dx}…\int v\frac{du}{dx}$$ All those ##(x)## bits look a bit messy to me:smile:

Another thing - I didn’t know you could evaluate the ##uv## part of IBP between the limits, as you have done in the square brackets. I haven’t seen that in any guide to integration, internet or book-based.

I thought the comment about initial terms vanishing was a maths/physics in joke, but there it is made real. Your method is a lot quicker and simpler than mine, which in turn was simpler than the integral calculator, which used u-sub in a very long-winded fashion. If you’d been my maths...

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 possible? If so, how is it done without dropping the bounds?

Stripping away the ##\lambda##, researching examples of the type: $$\int_0^\infty u^n e^{-u} du$$ I found that this is the gamma function for n+1: ##\Gamma (n+1)=n!## Obviously, if ##\lambda <0## in the original equation then we will no longer have ##e^{-\lambda u}## but ##e^{+\lambda u}## and...

Right, I tried to knock your question into bits, and accidentally found something called the Gamma function. First, the origin of the ##\lambda ^{n+1}## term. How did it get underneath, and why n+1? If I substitute ##u=\lambda x, x=\frac{u}{\lambda}, dx=\frac {du}{\lambda}##, I get...

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 back on for the top line, quoted in post #12: $$-(2ax^2e^{-2x/a})-(2a^2xe^{-2x/a}+a^3e^{-2x/a})$$ $$-2ax^2e^{-2x/a}-2a^2xe^{-2x/a}-a^3e^{-2x/a}$$...

Strewth, are we back to basics again? You may remember me from such threads as: https://www.physicsforums.com/threads/derivative-of-a-particles-energy.1015096/ Or, could have blanked from your mind the experience of dragging a biologist through simple algebra and calculus. I’ll have another...

The integral-calculator.com has it like this: $$-\frac{\left(2x^{2} + 2ax + a^{2}\right) \mathrm{e}^{-\frac{2x}{a}}}{4a}$$ Versus my answer: $$-\frac {(2x^2-2ax+a^2)e^{-2x/a}}{4a}$$ And when asked to do the definite integral, it’s correct: ##|C|^2\frac {a}{4}=1## according to the notes, (J...

Combining the result in post #4, which for some reason isn’t displaying, with Equation 1 of post #3: $$\frac{-ax^2 e^{-2x/a}}{2} - \frac {2a^2xe^{-2x/a} + a^3e^{-2x/a}}{4}$$ Which gives the extra minus sign that’s causing all the trouble.

Beyond my pay grade, I think ;)

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 {-ae^{-2x/a}}{2}## $$-a[\frac{-axe^{-2x/a}}{2}-(\frac {-a}{2}\frac{-ae^{-2x/a}}{2})]$$ $$ \frac{2a^2xe^{-2x/a} + a^3e^{-2x/a}{4} $$

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 {dv}{dx}=e^{-2x/a}, v=\frac {-ae^{-2x/a}}{2}## The IBP formula ##uv-\int v\frac{du}{dx}dx## gives me: $$\frac {-ax^2e^{-2x/a}}{2} - \int -ae^{-2x/a} x dx$$ Call...

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}## term, I went for integration by parts, a couple of times, first one being:##u=x^2, \frac {du}{dx}=2x, \frac {dv}{dx}=e^{-2x/a}, v=\frac {-ae^{-2x/a}}{2}## Now...