Answer
$10^4e^{0.0004 }$.
Work Step by Step
$$\int_{-\infty}^4e^{0.0001t}dt=\lim\limits_{R \to- \infty}\int_ R^4e^{0.0001t}dt \\=\lim\limits_{R \to -\infty} \frac{1}{0.0001}e^{0.0001t}|_R^4=\frac{1}{0.0001}\lim\limits_{R \to- \infty}(e^{0.0004 }-e^{0.0001R})=\frac{1}{0.0001}e^{0.0004 }.$$
Then the integral converges and its value is $10^4e^{0.0004 }$.
