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$$\int_0^\infty \frac{1}{t\sqrt t}e^{-1/t-pt} \, dt$$ $\operatorname{Re}(p)>0$

Michael Hardy
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math.
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  • If I am reading things right, the exponential term is constant, in which case there is a serious divergence problem for $\int_0^1$. – André Nicolas Jan 23 '16 at 05:48
  • my mistake sorry i've edited – math. Jan 23 '16 at 05:58
  • In general, for positive value of a and b, we have $$\int_0^\infty\exp\bigg(-a^2t-\dfrac{b^2}t\bigg)~\dfrac{dt}{t^{n+1}} ~=~ 2\bigg(\dfrac ab\bigg)^n~K_n(2ab),$$ see Bessel function for more information, which, for half-integer values of n, can be reduced to $$\dfrac{(ab)^m}{c^{2m+1}}\cdot\dfrac{\sqrt\pi}{e^{2ab}}\cdot \sum_{k=0}^m\dfrac{(m+k)!}{(m-k)!}\cdot\dfrac{(4ab)^{-k}}{k!},$$ with $m=|n|-\dfrac12$ and $c=a$ for $n<\dfrac12$ and $c=b$ for $n\ge\dfrac12.$ – Lucian Jan 23 '16 at 09:54
  • As usual, OEIS was essential in deducing the general formula. – Lucian Jan 23 '16 at 10:08

1 Answers1

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You are looking for the Laplace transform of $t^{-3/2} e^{-1/t}$. I would evaluate as follows: sub $t=1/u$ to get

$$\int_0^{\infty} \frac{du}{u^2} u^{3/2} e^{-u} e^{-p/u} = \int_{-\infty}^{\infty} dv \, e^{-(v^2 + p/v^2)} = e^{-2 \sqrt{p}} \int_{-\infty}^{\infty} dv \, e^{-\left (v-\sqrt{p}/v \right )^2 }$$

Let $v=p^{1/4} y$; then the integral is

$$p^{1/4} e^{-2 \sqrt{p}} \int_{-\infty}^{\infty} dv \, e^{-\sqrt{p} (y-1/y)^2}$$

Now, for well-behaved, even-valued functions $f$, i.e. integrable,

$$\int_{-\infty}^{\infty} dy \, f \left ( y-\frac1{y} \right ) = \int_{-\infty}^{\infty} dy \, f \left ( y \right ) $$

Thus, the integral is

$$p^{1/4} e^{-2 \sqrt{p}} \int_{-\infty}^{\infty} dv \, e^{-\sqrt{p} y^2} $$

Finally,

$$\int_0^{\infty} dt \, t^{-3/2} \, e^{-1/t} e^{-p t} = \sqrt{\pi} \, e^{-2 \sqrt{p}} $$

ADDENDUM

$$\int_{-\infty}^{\infty} dy \, f \left ( y-\frac1{y} \right ) = \int_0^{\infty} dy \, f \left ( y-\frac1{y} \right ) + \int_{-\infty}^0 dy \, f \left ( y-\frac1{y} \right )$$

$$\begin{align}\int_0^{\infty} dy \, f \left ( y-\frac1{y} \right ) &= \int_0^{1} dy \, f \left ( y-\frac1{y} \right ) + \int_1^{\infty} dy \, f \left ( y-\frac1{y} \right )\\ &= \int_1^{\infty} dy \, \left (1+\frac1{y^2} \right ) f \left ( y-\frac1{y} \right ) \\ &= \int_1^{\infty} d \left (y - \frac1{y} \right ) \, f \left ( y-\frac1{y} \right ) \\ &= \int_0^{\infty} dy \, f(y) \end{align}$$

Similar for $(-\infty,0]$.

Ron Gordon
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