Problem with Integration involving Logarithmic and Exponential Functions We have tried to determine the following integration problem. 
$\displaystyle\int_0^k \, dx \, \frac{x^2}{\sqrt{1 - (x/k)^2}} \, \ln\left[1 - e^{-x/t}\right] \quad k,\, t>0$ 
It would be very helpful if someone can help us to solve it. 
Also I would like to know, if we cannot solve a definite integration problem analytically, then what other techniques can possibly be used? Thanks! 
 A: Interesting integral.
For the moment, let's just rewrite it in a different way by simply unifying the denominator:
$$\int_0^k \frac{x^2}{\sqrt{\frac{k^2 - x^2}{k^2}}} \ln\left(1 - e^{-x/t}\right)\ \text{d}x$$
A little algebra leads you to
$$k\int_0^k \frac{x^2}{\sqrt{k^2 - x^2}} \ln\left(1 - e^{-x/t}\right)\ \text{d}x$$
Now let's perform a change of variable:
$$x = k\sin\theta ~~~~~ \text{d}x = k\cos\theta$$
Hence you get
$$k\int_0^{\pi/2} \frac{k^2\sin^2\theta}{\sqrt{k^2(1- \sin^2\theta)}} \ln\left(1 - e^{-k\sin\theta/t}\right)\ k\cos\theta \text{d}\theta$$
Which becomes easily
$$k^3\int_0^{\pi/2} \sin^2\theta \ln\left(1 - e^{-\alpha\sin\theta}\right)\ \text{d}\theta$$
Where $\alpha = k/t$.
First try, with approximation
Assuming the term $\alpha$ is large enough (not guaranteed, but as I said it's an approximation just to see what happens), we may think of a series expansion for the EXP up to the second order (cut):
$$e^{-\alpha\sin\theta} \approx 1 - \alpha\sin\theta$$
Hence
$$k^3\int_0^{\pi/2} \sin^2\theta \ln\left(1 - (1 - \alpha\sin\theta)\right)\ \text{d}\theta$$
$$k^3\int_0^{\pi/2} \sin^2\theta \ln\left(\alpha\sin\theta\right)\ \text{d}\theta$$
Due to the log properties, $\log(\alpha\sin\theta) = \log\alpha + \log\sin\theta$ and we spit into two integrals
$$(1) ~~~~~~~ k^3\int_0^{\pi/2} \sin^2\theta \ln(\alpha)\ \text{d}\theta = \frac{\pi}{4}k^3\ln(\alpha)$$
$$(2) ~~~~~~~ k^3\int_0^{\pi/2} \sin^2\theta \ln\left(\sin\theta\right)\ \text{d}\theta = -\frac{\pi}{8}k^3 (\ln(4)-1)$$
Hence in this case the solution would be
$$\frac{k^3\pi}{4}\left[\ln\left(\frac{k}{t}\right) - \frac{1}{2}\ln(4) + \frac{1}{2}\right]$$
Questions
Is that approximation valid? Well, yes and no. It all depends on the ratio $k/t$, and of course on how much we go on with the Taylor Series for the EXP.
Another way would be to use the Series for the entire LOG function, you may try.
Notice that in another very special case 
$$k = t = 1$$
You get the same answer 
$$-0.460655$$
Second order approximation
Just in case you want to go on, you'd have
$$e^{-\alpha\sin\theta} \approx 1 - \alpha\sin\theta + \frac{\alpha^2}{2}\sin^2\theta + \ldots$$
Hence the log would become
$$\ln\left(1 - \left(1 - \alpha\sin\theta + \frac{\alpha^2}{2}\sin^2\theta\right)\right) = \ln\left(\alpha\sin\theta\left(1 - \frac{\alpha}{2}\sin\theta\right)\right)$$
Which is again split into
$$\ln(\alpha\sin\theta) + \ln\left(1 - \frac{\alpha}{2}\sin\theta\right)$$
Integration of those terms would, this time, be not so easy. Indeed, except the first term which is easy the new second one is given in terms of Hypergeometric functions. You would have indeed at the second order
$$\frac{\pi}{8}\left(2 - \frac{9}{4}\ln(4) + \frac{8}{3\pi}\frac{k}{t} + 2\ln\left(\frac{k}{t}\right)\right) -\frac{1}{45} (k/t)^3 \, _3F_2\left(1,\frac{3}{2},3;\frac{5}{2},\frac{7}{2};\frac{(k/t)^2}{4}\right)+\frac{\pi  \left(\sqrt{4-(k/t)^2}-2\right)}{2 (k/t)^2}+\frac{1}{4} \pi  \log \left(\sqrt{4-(k/t)^2}+2\right)$$
This is what I could do for the moment.
Integrals of this type I guess they can be solved (if possible) only via numerical methods, approximations or other ways, but not analytically.
Anyway I'll think more about.
More on Hypergeometric Functions: https://en.wikipedia.org/wiki/Hypergeometric_function
