exact or numerical value of an improper integral

Question

i am dealing with an improper integral which has been arised in my research. i will be greatful if you have any idea about the numeric value of this integral. $$ \int_{0}^{\frac{1}{4}} \frac{u^{4} e^{8(1-u)}du}{\sqrt{e^{6}-16u^{2} e^{8(1-u)}}} $$ this integral has a singularity in $\frac{1}{4}$

Answer 1

The integrand is asymptotically equal to $$ \frac{e^3}{1024 \left(\frac{1}{4}-u\right)}-\frac{7 e^3}{768} $$ as $u$ approaches $\frac14$ from the left. Therefore the integral diverges.

Answer 2

This messy integral can be cleaned up quite a bit with minimal effort. We find:

$$\int_{0}^{\frac{1}{4}}\frac{u^{4}e^{8(1-u)}du}{\sqrt{e^{6}-16u^{2} e^{8(1-u)}}} = \int_{0}^{\frac{1}{4}}\frac{u^{4}e^{8}e^{-8u}du}{\sqrt{e^{6}-16u^{2} e^{8}e^{-8u}}}\\ = \int_{0}^{\frac{1}{4}}\frac{u^{4}e^{8}e^{-8u}du}{e^3\sqrt{1-16u^{2} e^{2}e^{-8u}}}\\ = e^3\int_{0}^{\frac{1}{4}}\frac{u^{4}e^{2}e^{-8u}du}{\sqrt{1-16u^{2} e^{2}e^{-8u}}}\\ = \frac{e^3}{1024}\int_{0}^{1}\frac{x^{4}e^{2}e^{-2x}dx}{\sqrt{1-x^{2} e^{2}e^{-2x}}}\\ = \frac{e^3}{1024}\int_{0}^{1}\frac{x^{4}e^{2(1-x)}dx}{\sqrt{1-x^{2} e^{2(1-x)}}}.$$

This last integral can more easily be seen to diverge because of the singularity at $x=1$, because near $x=1$ the integrand behaves as

$$\frac{x^{4}e^{2(1-x)}}{\sqrt{1-x^{2} e^{2(1-x)}}}\approx \frac{1}{x-1}+\frac73.$$