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Haha really? Did he also own a onesie? ;). Will keep an eye out for him!

Hey carrot, didn't get a chance to go through the mx2 thoroughly as I planned to, are you happy with how it is now?

Not personally, though I have probably been in the same room as him before. Why?

Look up the Gamma function and try to prove some of its basic properties, that is essentially what you are after.

Re: MX2 Integration Marathon Re: the restricted region of integration, keep in mind that you CAN integrate that function over any interval in R, it is just that your formula will only be valid in some places. Similarly, a slightly different formula would allow you to evaluate the integral over...

Re: MX2 Integration Marathon Within a restricted domain ;). Think about square roots carefully.

Not sure if that is a commonly used phrase? Perhaps "non-real", or "with non-zero imaginary part" are clearer. Up to you :).

Check 5 if you haven't already, in particular the meaning of the word "complex"... edit. and after that I'm pretty happy with the mc.

From what I can remember, an early MC question was wrong, and the last MC question didn't have any choices written yet lol. I think some subtleties regarding whether inequalities were strict or not were missed in the last question but that was from a cursory glance, I will give you something...

@Carrot. Didn't end up doing all of the exam on the bus, took a nap about 30 mins in and slept until sydney lol. But I think I found one or two issues. Will PM when I have looked more thoroughly.

Cosgrove is a bad lecturer but he was very very clever. He is better one-on-one.

Haha it was more to check for typos / errors / potentially confusing wording in questions, have been providing feedback on a couple of drafts so I know most of what's in it.

PM me the paper as it stands now so I can attempt it tomorrow p00nsticks :). (The MX2 one.)

Integrate log(t) from t=1 to t=x using the method used for this question, your answer will depend on the variable x. The derivative of this function is log(x) by the fundamental theorem of calculus, so adding an arbitrary constant to your answer gives you the indefinite integral of log x. (But...

Re: MX2 Integration Marathon Keep the trick in mind, it is a handy way to bypass some t-substitution bashes.

Re: MX2 Integration Marathon Alternatively alternatively, I=\int \frac{2}{(\sqrt{2}\cos(x+\pi/4))^2}\, dx=\int \sec^2(x+\pi/4)\,dx=\tan(x+\pi/4)+C

Re: MX2 Integration Marathon Easier.

Re: MX2 Integration Marathon Yeah, the initial substitution of mine was unnecessary of course. Just something I like doing, prefer to work with exponential functions rather than logarithmic ones.

Re: MX2 Integration Marathon There are many ways of justifying differentiating under the integral. Just use one that is general enough that it applies :). The one I use pretty much all the time is based on the dominated convergence theorem and suffices in most real-world applications.

Re: MX2 Integration Marathon Found a pretty slick way of doing this. $Let $G(s):=\int_0^1\frac{x^{s-1}-1}{\log(x)}\,dx=\int_{-\infty}^0 \frac{e^{sx}-e^x}{x}\,dx$ for real $s\geq 1$. \\$G(1)$ is trivially $0$. Now for $s\geq 1$ we have:\\ $G'(s)=\int_{-\infty}^0\frac{\partial }{\partial...