Is there a formula for the following integral? $$I(a,b)=\int_0^1 t^{3/2}(1t)^{1/2}\exp\left(\frac{a^2}{t}\frac{b^2}{1t} \right)dt$$ where $a,b$ are nonzero real numbers.

WolframAlpha isn't able to come up with anything. What is the motivation for this problem and what steps have you taken to evaluate it? – Gamma Function Jun 10 '13 at 08:16
2 Answers
I will derive the following result using the convolution theorem for Laplace Transforms:
$$I(a,b) = \int_0^1 dt \, t^{3/2} \, e^{a^2/t} \, (1t)^{1/2} \, e^{b^2/(1t)} = \frac{\sqrt{\pi}}{a} e^{(a+b)^2}$$
I assume $a$ and $b$ are $\gt 0$ for the derivation below, but you will see where the absolute values come from. It is very easy to check the correctness of this result with a few numerical examples in, say, Wolfram Alpha.
To begin, I refer you to the derivation of the following LT relation:
$$\int_0^{\infty} dt \, t^{3/2} \, e^{1/(4 t)}\, e^{s t} = 2 \sqrt{\pi} \, e^{\sqrt{s}}$$
We may rescale this to get
$$\int_0^{\infty} dt \, t^{3/2} \, e^{a^2/t}\, e^{s t} = \frac{\sqrt{\pi}}{a} \, e^{2 a\sqrt{s}}$$
Now the convolution theorem states that the convolution of the above
$$\int_0^1 dt \, t^{3/2} \, e^{a^2/t} \, (1t)^{3/2} \, e^{b^2/(1t)}$$
is equal to the inverse LT of the product of the individual LTs. This is easily expressed as follows:
$$\frac{1}{i 2 \pi} \int_{ci \infty}^{c+i \infty} ds \, e^s \frac{\pi}{a b} e^{2 (a+b) \sqrt{s}}$$
Note that this is being evaluated at $t=1$. And we of course know what the this integral evaluates to:
$$\int_0^1 dt \, t^{3/2} \, e^{a^2/t} \, (1t)^{3/2} \, e^{b^2/(1t)} = \sqrt{\pi} \left (\frac{1}{a} + \frac{1}{b} \right ) e^{(a+b)^2}$$
Of course, this is not the integral sought. But we may derive this integral by differentiating with respect to the parameter $b$:
$$\frac{\partial}{\partial b} \int_0^1 dt \, t^{3/2} \, e^{a^2/t} \, (1t)^{1/2} \, e^{b^2/(1t)} = 2 b \int_0^1 dt \, t^{3/2} \, e^{a^2/t} \, (1t)^{3/2} \, e^{b^2/(1t)} $$
which means we need to evaluate the following integral:
$$2 \sqrt{\pi} \int db \, b\, \left (\frac{1}{a} + \frac{1}{b} \right )\, e^{(a+b)^2} = \frac{2 \sqrt{\pi}}{a} \int db\, (a+b) e^{(a+b)^2} = \frac{\sqrt{\pi}}{a} e^{(a+b)^2} + C$$
Using the fact that the soughtafter integral goes to zero as $b \to \infty$, $C=0$ and we get
$$I(a,b) = \int_0^1 dt \, t^{3/2} \, e^{a^2/t} \, (1t)^{1/2} \, e^{b^2/(1t)} = \frac{\sqrt{\pi}}{a} e^{(a+b)^2}$$
BONUS
Of course, I could have considered the convolution between two different functions:
$$f(t) = t^{3/2} e^{a^2/t}$$
$$g(t) = t^{1/2} e^{b^2/t}$$
with corresponding LTs
$$\hat{f}(s) = \frac{\sqrt{\pi}}{a} \, e^{2 a\sqrt{s}}$$
$$\hat{g}(s) = \sqrt{\frac{\pi}{s}} \, e^{2 b\sqrt{s}}$$
(I will not derive the latter LT right now.) The convolution is then
$$\frac{\pi}{a} \frac{1}{i 2 \pi} \int_{ci \infty}^{c+i \infty} ds \, e^s \, s^{1/2} e^{2 (a+b) \sqrt{s}} = \frac{\sqrt{\pi}}{a} e^{(a+b)^2}$$
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Ron Gordon, thank you very much for your clear answer. The use of Laplace transform is very neat. – Kle Jun 11 '13 at 02:54