Integration by parts

We know from the product rule that $d(xy) = xdy+ydx$. Integrating it,
$$\int d(xy) = \int xdy + \int ydx$$
$$\therefore \int xdy = xy- \int ydx$$

This has applications in many different problems, and is known as integration by parts.

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Factorials, and the Gamma Function

What is the Gamma function? What does it have to do with the factorial of a number? And why does such a formula even work? The Factorial In our high school combinatorics course, we were introduced to the factorial of a natural number. We defined the factorial of a number $n$ to be the product of all the natural numbers till $n$. Or symbolically: $$n! = 1 . 2 . 3 ... (n-2) (n-1) n$$ This is a very useful function in Counting problems. This function also pops up in the exponential function: $$e^x = 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$ Setting this aside, we think: can we 'extend' this function to all real number? Surely, even if we were to come up with such a function, what would it mean to take this new 'factorial' of some real number? Well, let's take a look at one important property of the factorial function: $$n! = n(n-1)!$$ and this is true for all natural numbers $n$. So, even if we came up with this homologous function, we...

The $\operatorname{erf}$ function

Can a function be 'defined' as the anti-derivative of another function? $\operatorname{erf}$ is one such function. Even though it has various proper infinite series expansions (as in Wolfram MathWorld - Erf ), it is defined by mathematicians as such: $$\operatorname{erf}(z) \equiv \frac{2}{\sqrt{\pi}}\int_0^ze^{-t^2}dt$$ This function has extensive implications in statistics, and can be used to express the integral of $e^{-x^2}$ and $e^{x^2}$. See the post that tries to find integral of $e^{x^2}$ - we need not use integration by parts there, but rather can use the $\operatorname{erf}$ function. Source: Wolfram MathWorld

Why $y^2 = x^3 + 7$ has no integral solutions?

Consider the given equation... $$y^2-7=x^3=(x^{\frac{3}{2}})^2$$ $$y^2-(x^{\frac{3}{2}})^2=7$$ If $x^{\frac{3}{2}}$ is a decimal, then $y$ should also be some decimal to make the equation an integer, i.e., $7$. So $y$ will be a decimal in this case. But consider when $x^{\frac{3}{2}}$ is an integer. The only integers whose difference of squares is $7$, is $4$ and $3$ (why?). $$4^2 - 3^2 =7$$ But compare it to the equation $y^2-(x^{\frac{3}{2}})^2=7$. Then $$x^{\frac{3}{2}} = 3$$ $$x = 3^{\frac{2}{3}}$$ But here $x$ is a decimal! Therefore, at least one of the variables $x$ and $y$ is a decimal in the given equation, thus proving the fact that it has no integral solutions.

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