Present Value of Continuous Life Annuity - Derivation of Standard Integral Using Integration by Parts The standard integral expression for the present value of a continuous annuity on a life aged $x$ is:
$$\overline{a}_x = \int^\infty_0 v^t {}_t p_x dt $$
Notation:


*

*The symbol on the left hand side is the usual actuarial symbol for a continuous annuity contingent on a life aged $x$.

*$v=\frac{1}{1+i}$ is the accepted actuarial notation for the discount factor. It is the present value of one unit of money, one unit of time into the future. $i$ is the corresponding rate of (compound) interest.

*$v^t$ discounts by t units of time when applied as a factor.

*${}_tp_x = \mathbb P \left[ T_x \geqslant t \right]$ is the probability of the future lifetime of a life aged $x$ surviving to at least $t$.

*$T_x$ is the continuous random variable for the future lifetime of life aged $x$.


The method used to derive the integral expression follows the usual pattern for the expectation of a continuous random variable, in this case a function of $T_x$.
The derivation requires integration by parts, which is the reason for this post. Because the implementation of this technique is generally left to the reader as an exercise in the text books and courses I have seen, I wanted to implement it and post my workings ...
 A: The method I am accustomed to for stating the EPV (expected present value) as an integral starts with the expectation of a function of a continuous random variable, as here:
$$\overline{a}_x = \require{enclose} \mathbb{E}\bigl[ \overline a_\enclose{actuarial}{T_x} \bigr]$$
$T_x$ is the random variable for the future lifetime of $(x)$, with the distribution function $G(t)=\mathbb{P}[T_x<t]$, and PDF, $g(t)=\frac{d}{dt}G(t)$.
At this point I introduce some more standard actuarial notation, ${}_tq_x=G(t)$, with ${}_tp_x=1-G(t)$. I also note here the definition of the force of mortality as $\mu_{x+t}=-\frac{d}{dt} \ln(1-G(t))=\frac{g(t)}{1-G(t)}$. 
This last definition furnishes us with the standard actuarial expression for the PDF of $T_x$, namely $g(t)={}_tp_x \mu_{x+t}$. 
We note that $\overline {a}_\enclose{actuarial}{T_x}$ is a function of $T_x$. We further note a standard result of probability theory: a function $h$ of a continuous random variable $T$, which has a PDF of $g$, will have an expectation of $\mathbb E[h(T)] = \int h(t)g(t)dt$.
The pattern for constructing the integral then continues:
$$\mathbb{E}\bigl[ \overline a_\enclose{actuarial}{T_x} \bigr]=\int^\infty_0 \overline a_\enclose{actuarial}{_t} g(t) dt$$
$$=\int^\infty_0 \overline a_{\enclose{actuarial}t} {}_tp_x \mu_{x+t} dt$$
The trick then is to use integration by parts. Integration by parts tells us,
$$\int^\infty_0a(t)\frac{d}{dt}b(t)dt=\biggl[a(t)b(t)\biggr]^\infty_0 - \int^\infty_0b(t)\frac{d}{dt}a(t)dt$$
... where we choose, $a(t)=\overline a_{\enclose{actuarial}t}$; and, $\frac{d}{dt}b(t)={}_tp_x \mu_{x+t}$. By definition, we have,
$$\overline a_\enclose{actuarial}t=\int^t_0 e^{-\delta s}ds$$
$$=\frac{-1}{\delta}\biggl[ e^{-\delta s} \biggr]^t_0= \frac{-1}{\delta}\biggl[ e^{-\delta t} - 1\biggr] =\frac{1-v^t}{\delta}$$
[$v=e^{-\delta}$ is the actuarial discount factor used for calculating present values, which I introduced in the question. Note also that $\delta = \ln (1+i)$ is the force of interest.]
To proceed with integration by parts, we also require the derivative wrt $t$ of this expression:
$$\frac{d}{dt} \frac{1-v^t}{\delta}=v^t$$
This brings us to $b(t)$. Recall that ${}_tp_x \mu_{x+t}$ is the standard expression for the PDF of $T_x$, or $g(t)$ as it was introduced above. Therefore we have,
$$\frac{d}{dt}b(t)=g(t)=\frac{d}{dt}G(t) \implies b(t)=G(t)={}_tq_x$$
Now we proceed by turning our attention to the two terms on the right hand side of the integration by parts identity. 
$$\biggl[ a(t)b(t) \biggr]^\infty_0 = \biggl[ \frac{1-v^t}{\delta} \times {}_tq_x \biggr]^\infty_0 = \biggl[ \frac{1}{\delta} \times 1 - 0 \times 0 \biggr]=\frac{1}{\delta}$$
Note that the expressions in ${}_tq_x$ are evaluated as,
$${}_{\infty}q_x = \mathbb{P}[T_x<\infty]=1$$
and,
$${}_0q_x = \mathbb{P}[T_x<0]=0.$$
The second term from the identity is manipulated here,
$$\int^\infty_0 b(t) \frac{d}{dt} a(t) dt = \int^\infty_0 v^t {}_tq_x dt$$
$$= \int^\infty_0 v^t \left( 1-{}_tp_x \right) dt = \frac{1}{\delta} - \int^\infty_0 v^t {}_tp_x dt$$
Noting that, $\int^\infty_0 v^t dt = \frac{1}{\delta}$.
Then, putting all this together, we arrive at the required integral:
$$\overline a_x = \int^\infty_0 \overline a_{\enclose{actuarial}{T_x}} {}_tp_x \mu_{x+t} dt$$
$$= \frac{1}{\delta} - \frac{1}{\delta} + \int^\infty_0 v^t {}_tp_x dt$$
$$ =  \int^\infty_0 v^t {}_tp_x dt.$$
$\square$
