Integral exp(ax+H(x))

Question

How I can calculate

$$\int e^{(ax+bH(x))} dx $$

Where H is the Heaviside function.

Answer 1

$$H(x) = \left\{ \begin{array}{c} 1, x\ge 0 \\ 0, x <0 \end{array} \right.$$

$$I=\displaystyle \int e^{(ax+bH(x))} dx = I_1 + I_2$$

where $$I_1 = \int_{x<0} e^{(ax+bH(x)}) dx = \int_{x<0} e^{ax} dx$$ and

$$I_1 = \int_{x\ge 0} e^{(ax+bH(x)}) dx = e^b\int_{x\ge 0} e^{ax} dx.$$

$$I=\frac{1}{a} e^{ax} \left[ 1+ (e^b-1) H(x) \right] + \textrm{const.}$$