Recent content by PeroK

After the proposed u-substitution, the partial fractions step is trivial. In any case, it must be much simpler to do the u-substitution first.

:welcome:

What you've written isn't right at all. I understand what you are trying to do. I just don't see the point of making it look like a new magic formula that bears little relation to the product rule. The ##u## and ##v## substitutions look unnecessary to me. In a specific problem like this, we...

PS I don't understand why parts isn't taught this way. I never really got the point of that weird "uv" thing!

Parts is really the inverse product rule. Start with the product rule for differentiation: $$\frac{d}{dx}\big (f(x)g(x)\big ) = f'(x)g(x) + f(x)g'(x)$$Then integrate this equationbetween any two limits: $$\int_a^b \frac{d}{dx}\big (f(x)g(x)\big ) dx = \int_a^b f'(x)g(x) dx + \int_a^b f(x)g'(x)...

Integration by parts is valid for a definite integral. That allows you in this case to simplify the integral to a known standard form. Also, my style is to use parameters extensively, as I find I make fewer algebraic errors that way. In this case, I would take ##\lambda = \frac 2 a## and...

The equation I posted can be shown by induction on ##n##. What you stumbled on is an interesting idea for how to generalise the factorial beyond positive integers. Namely, the gamma function. First, we note that for positive integers ##n## we have: $$n! = \int_0^\infty u^n e^{-u} du$$Then, we...

:welcome:

Of the four users, only Bill has any authority on this site. I guess the mentors will make a decision when they see your post.

For in that metric tensor, What curvature may come, When we have shuffled off comoving time, Must give us pause. And rather bear the time we have, Than transform to others that we know not of.

Using the bounds on the definite integral is important in these cases, as it simplifies things considerably. For example, you could try to show (as an advanced exercise?!) that: $$\int_0^\infty x^n e^{-\lambda x} dx = \frac{n!}{\lambda^{n+1}} \ \ (\lambda > 0)$$Where ##n## is a positive...

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

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 step.

Why do you think that's wrong?

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 expansion always vanishes!