Integral of an involved exponential

singhofmpl
Messages
15
Reaction score
0

Homework Statement



I have problem solving an integral equation of the form


Homework Equations



\int_{0}^{\infty}(a\Lambda^2/(\Gamma(\gamma-\Lambda)^2))exp(\Lambda\gamma/(\Gamma(\gamma-\Lambda))-(\gamma-b)/c)d\gamma

The Attempt at a Solution


I have tried to solve it by substituting t=\frac{\Lambda\gamma}{\Gamma (\gamma-\Lambda)} and able to bring it to the following from:

\int_{0}^{\infty}-a\exp((c\Gamma t^2-(c\Lambda+\Lambda\Gamma-b\Gamma)t-b\Lambda)/(c(\Gamma t-\Lambda)))dt
but I'm not able to move further. Any suggestion to solve this integral will help me a lot.
 
Physics news on Phys.org
What is \Gamma? Is it a function or just a constant? If it is just a constant, you should be able to do a further substitution and complete the square inside the exponential to get it into the form \int_0^\infty \frac{e^{au^2}}{u}du. This has a solution in terms of the exponential integral (Ei).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top