Question: Verify:

$$y\left(\frac{1}{2}\right) = e^\frac{1}{2} \displaystyle \int\_\frac{1}{2}^{\infty} \frac{dt}{te^t} ≈ 0.9232.$$

I've written the integral as: 

$$\lim\_{x \to \infty} \displaystyle\int\_\frac{1}{2}^x \frac{dt}{te^t}$$

I presume I should use substitution to continue but not sure why or how.

We have gone over Trapezoidal and Simpson's Rule for numerical integration so I presume that's what I should use to solve it

Answer: 
Notice that

\begin{align\*}
\int\_{12}^{\infty} \frac{1}{te^t} \, dt &\le \sum\_{n = 12}^{\infty} \frac{1}{ne^n} \\
&\le \frac 1 {12e^{12}} \sum\_{n = 0}^{\infty} \frac 1 {e^n} \\
&< \frac{1}{e^{12}} \\
&< 0.000006
\end{align\*}

is already small enough that it won't affect your computation's first four decimal places. Now just compute $\int\_{1/2}^{12}$ using your favorite numerical method. (And by the way, this was a very crude method of estimate, so $12$ is a very loose upper bound.)
You substitute $u=\dfrac{1}{t}$ 

$dt = -\dfrac{du}{u^2}$

and get

$$\sqrt{e} \int\_2^0 \frac{(-1) u}{e^{1/u} u^2} \, du=\sqrt{e} \int\_0^2 \frac{e^{-1/u}}{u} \, du\approx 0.922911$$

[![enter image description here](https://i.stack.imgur.com/qc9TY.png)](https://i.stack.imgur.com/qc9TY.png)