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: 
Since the boundary does not enclose zero, change the integral variable $x=\frac{1}{t}$. Then the better method will be Gaussian quadratures to solve the integral numerically.

Assume $x=1/t$ then the integral becomes

$$\int\_0^2\dfrac{e^{-1/x}}{x}dx$$

then standardize it by defining $x=z+1$ the integral becomes

$$I = \int\_{-1}^1F(z)dz=\int\_{-1}^1\dfrac{e^{-1/(z+1)}}{z+1}dz$$

you may use any Gauss-Legendre formula via

$$I=\sum\_{i=0}^nw\_iF(z\_i)$$

For four-point formula when $n=3$, 

$$I=0.5590$$

if you multiply it by $e^{1/2}$, your final value is 0.9216.
Use the substitution 

$$
t = \frac{1}{2}+\frac{x}{1-x}
$$

Note that when $x\to 1$ then $t\to\infty$. Your integral then becomes

$$
y(1/2) = -2e\int\_0^{1}{\rm d}x~\frac{e^{1/(x-1)}}{x^2-1}
$$

The result is 

$$
y(1/2)\approx 0.922911
$$