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May I ask how to do the following integration?

$$\int_0^\infty \frac{e^{-(\pi n^{2}/x) -(\pi t^2 x)}}{\sqrt{x}} dx $$

where $t>0$, $n$ a positive integer.

This came up on page 32 (image) of Titchmarsh's book, The Theory of the Riemann Zeta-Function. Specifically for the sum involving $b_n$, I am wondering how to

multiply by $e^{-\pi t^{2} x}$ and integrate over $(0,\infty)$

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I hope you don't mind, I added an image of the relevant page and rephrased your question slightly. Please feel free to make any changes you see fit. –  Zev Chonoles Jul 9 '12 at 5:09
    
Thanks for the editing, this looks great. –  ericc Jul 9 '12 at 6:03

1 Answer 1

up vote 6 down vote accepted

$$\begin{eqnarray*} \int_0^\infty \frac{dx}{\sqrt{x}}\, \exp\left(-\frac{\pi n^2}{x} - \pi t^2 x\right) &=& 2\sqrt{\frac{n}{t}} \int_0^\infty dz\, \exp\left(-\pi n t(z^2 + z^{-2})\right) \hspace{5ex} (\textrm{let } x=z^2 n/t) \\ &=& \sqrt{\frac{n}{t}} \int_{-\infty}^\infty ds\, e^{s/2} \exp\left(-2 \pi n t \cosh s\right) \hspace{5ex} (\textrm{let } z=e^{s/2}) \\ &=& 2\sqrt{\frac{n}{t}} \int_0^\infty ds\, \cosh\left(\frac{s}{2}\right) \exp\left(-2 \pi n t \cosh s\right) \\ &=& 2\sqrt{\frac{n}{t}} K_{\frac{1}{2}} (2\pi n t) \hspace{5ex} (\textrm{modified Bessel function, 2nd kind}) \\ &=& \frac{e^{-2\pi n t}}{t} \end{eqnarray*}$$

Addendum: An approach not involving special functions. $$\begin{eqnarray*} \int_0^\infty \frac{dx}{\sqrt{x}}\, \exp\left(-\frac{\pi n^2}{x} - \pi t^2 x\right) &=& 2\sqrt{\frac{n}{t}} \int_0^\infty dz\, \exp\left(-\pi n t(z^2 + z^{-2})\right) \hspace{5ex} (\textrm{as before}) \\ &=& 2\sqrt{\frac{n}{t}} \int_0^\infty dz\, \exp\left(-\pi n t(z-z^{-1})^2 - 2\pi n t\right) \\ &=& \sqrt{\frac{n}{t}} e^{-2\pi n t} \int_{-\infty}^\infty du\, \left(1+\frac{u}{\sqrt{u^2+4}}\right) e^{-\pi n t u^2} \hspace{4ex} (z-z^{-1}=u) \\ &=& \frac{e^{-2\pi n t}}{t} \hspace{5ex} (\textrm{odd integral vanishes; Gaussian left over}) \end{eqnarray*}$$

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Nice work here !(+1) –  Chris's sis Oct 15 '12 at 22:59
    
@Chris'ssister: Cheers! –  user26872 Oct 16 '12 at 0:08

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