Numerical integration with infinity as upper bound 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
 A: 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.)
A: 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.
A: You substitute $u=\dfrac{1}{t}$ 
$dt = -\dfrac{du}{u^2}$
and get
$$\sqrt{e} \int_2^0 \frac{(-1) u}{e^{1/u} u^2} \, du=\sqrt{e} \int_0^2 \frac{e^{-1/u}}{u} \, du\approx 0.922911$$

A: 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 
$$
y(1/2)\approx 0.922911
$$
A: 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:
$$ \mathfrak{I} \approx \frac{1}{100}\left[2+\sum_{k=1}^{99}\frac{\tfrac{k}{100}}{\tfrac{1}{4}-\log\tfrac{k}{100}}\right] \approx 0.923. $$
