Guineafowl's latest activity

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...

Parts is really the inverse product rule. Start with the product rule for differentiation: $$\frac{d}{dx}\big (f(x)g(x)\big ) =...

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...

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...

Integration by parts is valid for a definite integral. That allows you in this case to simplify the integral to a known standard form...

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 ;)