Integration by parts is a technique used in calculus to integrate products of functions, especially when both parts of the product depend on the variable of integration. $ \int u \: dv = uv - \int v \: du $ * $u$: A function we will differentiate during the procedure ($u \to du$). * $dv$: A function we will integrate during the procedure ($dv \to v$). With this formula we can transform the integral of a product $\int u \: dv$ into a combination of product and another ==hopefully simpler integral== $\int v \: du$. **Example:** Let $X$ be a [[Random Variable]] an from [[Exponential Distribution]], with the following [[Probability Density Function]]. $ f_X(x)=\lambda e^{-\lambda x}, \quad x \ge 0, \lambda >0.$ To obtain the expectation we will use integration by parts. $ \mathbb E[X] =\int_0^\infty \underbrace{x}_{u}* \underbrace{\lambda e^{-\lambda t} \, dx}_{dv} $ Terms: - $u=x$ - $du = dx$ (derivative of $u$) - $dv = \lambda e^{-\lambda t}dt$ - $v= e^{-\lambda t}$ (integrated $dv$) $ \begin{align} \mathbb E[X] &= \int_0^\infty x* \lambda e^{-\lambda x} dx \tag{1}\\[4pt] &= \Big[x* -e^{-\lambda x} \Big ]_0^\infty - \int_0^\infty -e^{-\lambda x} dx \tag{2}\\[4pt] &= \underbrace{\Big[x* -e^{-\lambda x} \Big ]_0^\infty}_{=0} +\Big[\frac{1}{\lambda} e^{-\lambda t} \Big ]_0^\infty \tag{3}\\[4pt] &=1/\lambda \tag{4} \end{align} $ (3) Evaluating the term at $[0, \infty]$ both boundaries returns $0$, which make the whole terms $0$. * When $x = \infty$ then $xe^{-\lambda x} = 0$ because the exponential decay is stronger than the linear growth in $x$. * When $x=0$ then $xe^{-\lambda x} \to 0$.