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)
We have $$\mathfrak{I}=e^{1/2}\int\_{1/2}^{+\infty}\frac{dx}{x e^x}\,dx = \int\_{0}^{+\infty}\frac{dt}{\left(t+\tfrac{1}{2}\right)e^t}=\int\_{0}^{1}\frac{du}{\tfrac{1}{2}-\log u}=\int\_{0}^{1}\frac{v}{\tfrac{1}{4}-\log v}\,dv$$
where the last form is well-suited for standard techniques: the integration range is a bounded interval and the integrand function has bounded derivatives over the closed interval. For instance, through the trapezoidal rule:
$$ \mathfrak{I} \approx \frac{1}{100}\left[2+\sum\_{k=1}^{99}\frac{\tfrac{k}{100}}{\tfrac{1}{4}-\log\tfrac{k}{100}}\right] \approx 0.923. $$