The Proof is Trivial!

Another solution to 295:

$\displaystyle I = \int_{-\infty}^{\infty} e^{-x^2} \ dx = 2 \int_0^{\infty} e^{-x^2} \ dx \overset{x^2 \mapsto x}= \int_0^{\infty} e^{-x} x^{-\frac{1}{2}} \ dx = \Gamma \left(\tfrac{1}{2}\right)$

Now, as $\Gamma (z) = \int_0^{\infty} x^{z-1} e^{-x} \ dx$ and $\text{B} (x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x+y)}$, we have:

$\text{B}\left(\frac{1}{2}, \frac{1}{2}\right) = \dfrac{\Gamma^2 \left(\frac{1}{2}\right)}{\Gamma (1)} = \Gamma^2 \left(\frac{1}{2}\right) = I^2$

$\displaystyle \text{B}\left(\tfrac{1}{2}, \tfrac{1}{2}\right) = \int_0^1 x^{-\frac{1}{2}} (1-x)^{-\frac{1}{2}} \ dx \overset{x^{\frac{1}{2}}\mapsto x}= 2\arcsin{x}\bigg|_0^1 = \pi$

Therefore $I^2 = \pi \implies I = \sqrt{\pi}$

Is there a method of solving this integral using just elementary methods (i.e. no double integrals or change of coordinate systems).

Is there a method of solving this integral using just elementary methods (i.e. no double integrals or change of coordinate systems).

Problem 365***(revisited)

Find $\displaystyle \int_{0}^{\infty} \frac{x^{a}}{(b+x^{c})^{d}}dx$, when $\displaystyle a \ge 0, b>0, c>0, d > \frac{a+1}{c}$. We also throw out the condition $d \in \mathbb{Z}$.

It's still a really good question, there're some quite complex in-depth ways of doing it which I don't understand at all, would love to see some people post them. What was your method?

This looks so hard, not even gonna lie

This looks so hard, not even gonna lie

This looks so hard, not even gonna lie

Hard is good, makes it even more satisfying when you complete a problem I'm sure there'll be something on here you could do, I manage to find the odd thing occasionally

Which question?

Good! I was aiming for that first impression.

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Hard is good, makes it even more satisfying when you complete a problem I'm sure there'll be something on here you could do, I manage to find the odd thing occasionally

Most of the easier questions on here are the ones that I have posted. Some would say that I have diluted the thread but in my mind, I'm making it more accessible

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Very much so, could you quickly summarise what neutrices are? It seems counterintuitive that a function can have a defined derivative at a point where it is not defined.

If you are having a boring Wednesday night, then play with $n!^{(k)} + a = m^2$ to find solutions $(n,m)$ for $k,a \in \mathbb{N}$.

For example, $n!! + 2 = m^2$.

I used the polar coordinates method as well

I haven't seen any differential equations on this thread so here is one.

Problem 398***
Find the general solution to the system of differential equations.
$y'_1= 4y_1 - y_3$
$y'_2 = 2y_1 + 2y_2 - y_3$
$y'_3 = 3y_1 + y_2$

Problem 396**

Hopefully I'll solve Mladenov's problems at some point, but in the hiatus here's another integral:

$\displaystyle \int_0^{\frac{\pi}{2}}\frac{\log(\sec(x))}{\tan(x)}dx.$

Problem 397**

$\displaystyle \int_0^{\frac{\pi}{2}}\frac{\log(\sec(x))}{\tan^2(x)}dx.$

Solution 397
Seemed a bit too simple so most probably I made mistake (I hope log(x)=ln(x))
$u=tanx$
$\displaystyle \int \frac{\log(\sec(x))}{\tan^2(x)}dx=\dfrac{1}{2} \int \dfrac{log(u^2+1)}{u^2}du \ _{\rightarrow}^{IBP} \ \ \ \ \dfrac{1}{2} \left( 2arctanu-\dfrac{log(u^2+1)}{u} \right)$
$\therefore \displaystyle \int_0^{\frac{\pi}{2}}\frac{\log(\sec(x))}{\tan(x)}dx= \dfrac{1}{2} \left[arctan(tan(x))-\dfrac{sec^2x}{tanx} \right]_0^{\frac{\pi}{2}}$
Assuming I've made no mistakes (surprising as it's midnight), I'm stuck as I'm not sure how to evaluate $arctan(tan(\frac{\pi}{2}))$ although googling suggests it might be 0. Not sure though.

Well, $\arctan{\tan{x}} = x$ as you're integrating over $\left[0,\frac{\pi}{2}\right]$ but you've done the substitution incorrectly, you've forgotten to make an appropriate substitution for $dx$. Your integral is actually transformed to:

$\displaystyle \int_0^{\frac{\pi}{2}} \frac{\ln{\sec{x}}}{\tan^2 x} \ dx \overset{u=\tan{x}}= \frac{1}{2}\int_0^{\infty} \frac{\ln{(u^2 + 1)}}{u^2 (u^2 + 1)} \ du$.

Depends what you class as an elementary method. The function has no antiderivative, if that's what you mean.

To be fair, since I only started learning differential just recently, I had to get some help with this question.

$A=\begin{pmatrix}4&0&-1\\2&2&-1\\3&1&0\end{pmatrix}$ which has a defined polynomial $\chi_A=(X-2)^3$. Which also has a Jordan form $\begin{pmatrix}2&0&0\\1&2&0\\0&1&2\end{pmatrix}$ which will give a system $X'=CJC^{-1}X$, which should be solved by the change of variables $Y=C^{-1}X$.

But, sadly I don't know how Maybe someone better can take it from here.

Can you please tell me where you got this question? I want to read more into these types of questions.