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: 
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)
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 

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