2
$\begingroup$

I am trying to evaluate definite integral:

$$\frac{2}{876} \int_{0}^{T} \left[ x^{-3/8} (1+b\,x)^{3/4} (1+ab\,x)^{-3} e^{(c\,x^2)}+\log{(5/4)}\right]\,\mathrm dx$$

$a$, $b$, $c$ are all positive ($a=3.8/3777$, $b=3777$, $c=4.8\times 10^{-3}$). I have plotted the whole integrand. In the plot $\tau$ corresponds to $x$ in the integral. The parameter $\eta$ can be ignored. I have tried the Leibniz method (also called Feynman's integration trick) but without any success. Maybe with a clever change of variable it can be expressed in terms of (combinations of) special functions... or at least get some bounds? Actually I have a different way to estimate the integral, since it's a physics problem and I can guess the (upper limit) of it. T should be $T\sim 57$ (months), I hope. Thanks.

enter image description here

$\endgroup$
  • $\begingroup$ I made a significant improvement from a formal point of view. $\endgroup$ – Claude Leibovici May 11 '20 at 4:10
1
$\begingroup$

"abandon hope all ye who enter here" as wrote Dante Alighieri in "The Divine Comedy"

Using your numbers, you have $$f(x)= \frac{1}{438} \left(\frac{ (1+3777 x)^{3/4}}{x^{3/8} \left(1+\frac{19 }{5}x\right)^3}\,e^{\frac{3 x^2}{625}}+\log \left(\frac{5}{4}\right)\right)$$ and it seems to me that you want to find the values of $T$ such that $$F(T)=\int_0^T f(x) \,dx = \eta $$ There is absolutely no hope to get a closed form formula for the result and, for whatever you could need to do, all the work needs numerical methods.

What is interesting is to look at the plot of $G(T)=\log(F(T))$ as a function of $T$; there are two parts : an almost horizontal part as long as $\eta < 1$ and for $\eta >1$ $G(T)$ exhibits an almost parabolic shape.

For solving the equation, Newton method is very effective for finding the zero of the equation $$G(T)-\log(\eta)=0$$ Using in particular the fundamental theorem of calulus, the iterates will be given by $$T_{n+1}=T_n-\frac{F(T_n)}{f(T_n)}\log \left(\frac{F(T_n)}{\eta }\right)$$

Let us try with $T_0=123$ and $\eta=1$. The iterates will be $$\left( \begin{array}{cc} n & T_n \\ 0 & 123.000 \\ 1 & 74.4050 \\ 2 & 56.6963 \\ 3 & 53.1111 \\ 4 & 52.8262 \\ 5 & 52.8235 \end{array} \right)$$ which is quite fast in spite of a quite poor estimate.

In fact it seems that a rather good estimate could be $$T(\eta)=52.8235 + a \big[\log(\eta)\big]^b$$ A quick a dirty non linear regression gives (with $R^2=0.999906$) $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 9.08300 & 0.03959 & \{9.00532,9.16069\} \\ b & 0.54957 & 0.00055 & \{0.54848,0.55066\} \\ \end{array}$$

  • For $\eta=10^{100}$, the estimate is $233.304$ while the solution is $227.851$
  • For $\eta=10^{1000}$, the estimate is $692.559$ while the solution is $696.057$
$\endgroup$
0
$\begingroup$

Thanks a lot for the insight. The integral does have an analytical solution, but it is not very helpful, because I need to evaluate "chunks" of it. The solution (for completeness) is

$$\begin{align} f(T) & = \sum_{n=0}^\infty\frac{c^n T^{2n+5/8}}{n!}\frac{\Gamma(2n+5/8)}{\Gamma(2n+13/8)}F_1\left(2n+\frac{5}{8},-\frac{3}{4},3,2n+\frac{13}{8};-bT,-aT\right)\nonumber\\ & = \sum_{n=0}^\infty \frac{c^n}{2n+5/8}\frac{T^{2n+5/8}}{n!}F_1\left(2n+\frac{5}{8},-\frac{3}{4},3,2n+\frac{13}{8};-bT,-aT\right) \end{align} $$

This was obtained by expanding the exponential function as a power series, changing the variables in order for the limits of the integral to be 0 and 1 and using the integral representation of the first Appell hypergeometric series, with the note that while these expression hold only for specific arguments, they can be analytically extended to larger domains.

However, I agree that one needs to evaluate it numerically to obtain specific results at given times. Thanks!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.