find normalization constant a for $f(x)=a x\sin(x) e^{-x}$

Question

The continuous random variable X has the joint probability density

$f(x)=a x \sin(x) e^{-x}$ for

$0<x<\infty$ and 0 otherwise.

Find the normalization constant a.

I am not sure how to find the normalization constant.

Answer 1

You know that $$ \int_{0}^{+\infty}a\sin\left(x\right)e^{-x}\text{d}x=1 $$ Hence you can deduce $a$. If you have difficulties computing the improper integral, I stringly recommend you to see what it looks like in a complex form. $$ \int_{0}^{+\infty}\sin\left(x\right)e^{-x}\text{d}x=\text{Im}\left(\int_{0}^{+\infty}e^{x\left(i-1\right)}\text{d}x\right)=\text{Im}\left(\left[\frac{e^{x\left(i-1\right)}}{i-1}\right]^{+\infty}_{0}\right)=\text{Im}\left(\frac{1}{1-i}\right) $$ Then $$\frac{1}{1-i}=\frac{1+i}{2} $$ You deduce $$ \int_{0}^{+\infty}\sin\left(x\right)e^{-x}\text{d}x=\frac{1}{2}$$ And finally $$ a=2 $$

Answer 2

Since $f(x)$ is a probability density function:

$\int_{0}^{+\infty}ax\sin\left(x\right)e^{-x}\text{d}x=1$

Therefore, $a =\frac{1}{\int_{0}^{+\infty}x\sin\left(x\right)e^{-x}\text{d}x}$

And you are lucky because, $\int_{0}^{+\infty}\sin\left(x\right)e^{-x}\text{d}x =\int_{0}^{+\infty}x\sin\left(x\right)e^{-x}\text{d}x = 0.5$

Therefore, $a=2$