Hazard function and survival analysis I have a function $\lambda(t)$ which returns the instant probability of dying at a time $t$. I'd like to compute the function $F(t)$ which returns the probability of being dead at a certain time $t$.
Using some reliability theory, I have obtained the following relation:
$$
F(t) = 1-e^{- \int_{0}^{t} \lambda(u) \,du }
$$
But when I do a numerical application I obtain strange results:
Here is the shape of the function $\lambda(t)$:

It can be seen that just after 60 seconds, the probability of dying is above $0.2$. But then when I try to compute the probability of beeing dead at $t=100s$, I obtain:
$$
F(t) = 1-e^{- \int_{0}^{100} \lambda(u) \,du }=0.051
$$
What am I doing wrong?
 A: Answer
Given that your $\lambda(t)$ actually represents the instantaneous probability to die at time $t$ (the hazard), your function $F(t)$ computes the probability to die before a certain time $t$. In other words, $F(t)$ is the cumulative distribution function (CDF) of the time of death.
Following the discussion in the comments, your function represents the probability density function (pdf) $f(t)$ of time of death. From there you can compute the CDF as
$$
F(t) = \int_0^T f(u) du
$$
with $T$ either infinity or the largest possible time.
Some Background on Survival Models / Distributions
The following derivations are largely based on this material
Survival models can be based on the hazard rate $\lambda(t)$ (the instantaneous probability to die at time $t$) or on the distribution of times of death $f(t)$ (which is the pdf of the random variable time of death).
\begin{align}
 \text{Hazard function}   && \lambda(t) &= \underset{\mathrm{d}t \rightarrow 0}{\lim} \frac{P(t \leq T \leq t + \mathrm{d}t | T\geq t)}{\mathrm{d}t} = \frac{f(t)}{S(t)} = - \frac{\mathrm{d}}{\mathrm{d}t}\ln \big[S(t)\big] \nonumber \\
 \text{Cumulative Hazard} && M(t) &= \int_0^t \lambda(x) \mathrm{d}x \label{eq:cum_hazard} \\
 \text{pdf}               && f(t) &= \lambda(t) S(t) \nonumber \\
 \text{cdf}               && F(t) &= P(T < t) = \int_{0}^t f(x)  \mathrm{d}x \nonumber \\
 \text{Survival function} && S(t) &= 1- F(t) = \int_t^{\omega}f(x) \mathrm{d}x = e^{-M(t)} \label{eq:surv}\\
 \text{Life Expectancy}   && e(t) &= \frac{1}{S(t)} \int_t^\omega S(x) \mathrm{d}x \quad \text{with } e(0) = \int_0^\omega S(x) \mathrm{d}x \quad \text{since} \quad S(0) = 1 \label{eq:lexp}
\end{align}
where $\omega$ denotes the highest age.
A: If your plot is intended to be the probability density function for an event over time, $f(t)$ in the terminology of the answer from @Martin Georg Haas, then you technically have an improper distribution as it does not integrate to 1 over its support. Your numerical evaluation of the integral, consistent with $F(\infty) = 0.051$, agrees well with the area of a roughly corresponding isosceles triangle having height 0.2 and base 0.5.
If that's what you intend, you have a cure model in which about 95% of the cases never experience the event: $$S(\infty)= 1 - 0.051=0.949.$$ If that matches your data, you haven't done anything "wrong."
If that's not what you intend and all individuals eventually experience the event, then you have done something wrong and you need to scale your $f(t)$ by a factor of about 20 so that it integrates to $F(\infty)=1$.
