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Here we consider the function $$I(t)=\frac{1}{a\int_{-\infty}^{t}e^{-\int_r^t \lambda (\tau) \ d\tau}dr}$$ And I am trying to find a function $\lambda(\tau)$ such that $I(t)$ converges. Here we take $a$ As constant. I have tried taking $\lambda (\tau)=\frac{1}{\tau}$. I obtained an integral whose solution did not converge. Any help would be greatly appreciated!

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    $\begingroup$ $I(t)$ converges when $t$ goes to a certain limit, or just $I(t)$ is well defined for every $t$? $\endgroup$ – LL 3.14 May 2 '20 at 0:02
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    $\begingroup$ $I(t)$ is a function of $t$. It is a variable which changes. So I am trying to evaluate the integral so I can plot the function $I(t)$ against $t$. When evaluated the function should be entirely dependent on $t$ $\endgroup$ – MATHBOI May 2 '20 at 0:18
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If you're free to choose $\lambda$ to be whatever, try a simple power law.

$$ \lambda (\tau) = \tau^k\\ I(t)=\frac{1}{a\int_{-\infty}^{t}e^{-\int_r^t \tau^k \ d\tau}dr}\\ I(t)=\frac{1}{a\int_{-\infty}^{t}e^{-\frac{1}{k+1} (t^{k+1}-r^{k+1})}dr}\\ I(t)=\frac{1}{a\int_{-\infty}^{t}e^{-\frac{1}{k+1} t^{k+1}} e^{-\frac{r^{k+1}}{k+1}}dr}\\ I(t)=\frac{1}{a e^{-\frac{1}{k+1} t^{k+1}} \int_{-\infty}^{t} e^{-\frac{r^{k+1}}{k+1}}dr}\\ I(t)=\frac{e^{\frac{1}{k+1} t^{k+1}} }{a \int_{-\infty}^{t} e^{-\frac{r^{k+1}}{k+1}}dr}\\ $$

We need to make sure this all makes sense for given $k$. Everything converging and all. This is where picking $k$ comes in.

If you use $k=1$, the denominator looks like integrating a Gaussian and the numerator is sign flipped from a Gaussian. It blows up real fast but that might be okay for you.

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