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I'm trying to find the result of the following definite integral

$$ \int_0^1 \frac{d t}{(t(1-t))^{3/2}}\, \exp\left(-\frac{a}{t}-\frac{b}{1-t}\right)$$ for positive $a$, $b$.

Any hints or possible methods to attack the problem are welcome.

user72829
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    See this: http://math.stackexchange.com/questions/1148493/a-difficult-one-variable-exponential-integral/1148627#1148627. Note that this is a convolution of two functions whose inverse Laplace transforms are known. The result you want is the derivative of the answer in the link with respect to $a$ and $b$. – Ron Gordon Dec 01 '15 at 14:05
  • from where do you got this integral? – Dr. Sonnhard Graubner Dec 01 '15 at 14:19

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OK, let's quickly work this one out based on the linked-to answer. Note that

$$\int_0^1 dt \frac1{(t (1-t))^{3/2}} e^{-\left (\frac{a}{t} + \frac{b}{1-t} \right )} = \frac{\partial^2}{\partial a \, \partial b}\int_0^1 dt \frac1{(t (1-t))^{1/2}} e^{-\left (\frac{a}{t} + \frac{b}{1-t} \right )} = \pi \frac{\partial^2}{\partial a \, \partial b} \operatorname{erfc}{\left ( \sqrt{a} + \sqrt{b}\right )}=\sqrt{\pi} \left (\frac1{\sqrt{a}} + \frac1{\sqrt{b}}\right )\exp\left(-\left(\sqrt{a}+\sqrt{b}\right)^2\right)$$

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Ron Gordon
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$a, b$ are positive and the range of integration is $[0, 1]$. What about a Taylor series for the Exponential?

$$e^{\left(-\frac{a}{t} - \frac{b}{1-t}\right)} = e^{\frac{t(a-b) -a}{t(1-t)}}$$

From this, using Taylor,

$$e^{\frac{t(a-b) -a}{t(1-t)}} = \sum_{k = 0}^{+\infty} \frac{1}{k!}\left(\frac{t(a-b) - a}{t(1-t)} \right)^k$$

Substituting into the integral you get:

$$\sum_{k = 0}^{+\infty} \frac{1}{k!}\int_0^1 \frac{\text{d}t}{(t(1-t))^{\frac{3}{2}}}\left(\frac{t(a-b) - a}{t(1-t)} \right)^k = \sum_{k = 0}^{+\infty} \frac{1}{k!}\int_0^1 \frac{(t(a-b) - a)^k}{(t(1-t))^{\frac{3}{2} + k}} \text{d} t$$

Then one should see what's next..