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I've been trying to find a closed form expression/series expansion for the following integral without success:

$$F(a,b)=\int_{\epsilon-i\infty}^{\epsilon+i\infty} e^{az+b^2z^2}\Gamma(z)\Gamma(1-z)dz$$

where $a, b, \epsilon>0$. Any input is greatly appreciated!

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Maybe you can start with Gamma function reflection formula $\Gamma(z)\Gamma(1-z)=\pi \csc \pi z$. –  i707107 Mar 3 '13 at 5:20
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what are the signs of a and b? maybe you could use the reflection formula as suggested and possibly use contour integration to get a sum over the poles of $\sin \pi z $. This of course would require the integrand to decay, which is why I asked about a and b... –  Tom Dickens Mar 3 '13 at 5:31
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I meant to say the poles at the zeros of $\sin \pi z$ ! –  Tom Dickens Mar 3 '13 at 6:06
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I have not had a chance to work it all out, but you could try writing the sin in terms of exponentials and then expanding the result in a series in $\exp(\pi x)$; then you would simply have Gaussian integrals. You would need to take care to split the range at the real axis. (Should be symmetric about it anyway.) –  Tom Dickens Mar 4 '13 at 4:14
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I think the thing to do is split the csc as (csc x - 1/x) + 1/x, then the integrand with the term in parentheses is not singular and can be pushed to the axis. The pole term can be done analytically as a principal value. I believe the first term will give a series that can be integrated term by term. I have been playing with this and will get you an answer in a couple of days... i'm busy...The pole term gives $i \pi$ + an error function. –  Tom Dickens Mar 7 '13 at 4:10

1 Answer 1

Hi dp,

Using the approach I mentioned above, that is splitting the $\csc(\pi z)$ term into its pole part $\frac{1}{\pi z}$ and the rest, I get an exact answer for the pole part and a series for the rest. The pole term yields a principal value integral that can be computed analytically, plus the contribution from integrating around an infinitesimal semi-circle, giving

$$ I_1 = i\pi + i\pi\ \mbox{erf} \left( \frac{a}{2 b} \right), $$

where $\mbox{erf}()$ is the error function.

The remaining contribution can be evaluated as an asymptotic series for large $b$. We write $$ \csc(\pi u ) - \frac{1}{\pi u} = \sum_{m=1} c_{2 m-1} ( u)^{2m -1} ,$$ which has a radius of convergence of $\vert u\vert \le 1.$ Therefore, if we substitute this in the integrand, the resulting series will be an asymptotic series for large $b$, since the series has a finite radius of convergence. Since this integral has no singularity on the real axis (please note, I have changed variables using $z = i u$, so the integration is along the real axis), we may take $\epsilon=0$ and write

$$ I_2 \approx i\pi \int_{-\infty}^{\infty} \exp(i a u) \exp(-b^2 u^2) \sum_{m=1} c_{2m-1} i^{2m-1} u^{2m-1} . $$

For clarity write $b^2 \rightarrow g.$ We use $u^{2m} = \left( -\frac{\partial^m}{\partial g^m}\right) \exp(-g u^2).$ Reversing the order of the sum and integral and performing the integral gives

$$ I_2 \approx i \pi \sum_{m=1} c_{2m-1} \frac{\partial^m}{\partial g^m} \pi\ \mbox{erf}\left( \frac{a}{2 \sqrt{g}} \right) , $$

where $g = b^2$.

May I ask, from what problem does this integral arise? Is it from an inverse Mellin transform used to perform some other integral? Do you expect the parameter $b$ to be significantly larger than one?

I add this - numerically integrating your original integral in Mathematica for $a=2, b=2$ gives $I \approx 4.56 i$ while the main term above gives $I_1 \approx 4.78 i $. For $a=2, b=4$ the results are $3.97 i, 4.01 i.$

I will include additional details here later, with some results for the second term $I_2.$

EDIT - For small $b$, one can evaluate the integral by expanding the term $e^{b^2 z^2}$ in a series. The term coming from the principal value integral is kept the same as shown above, while the other integral can be written as

$$ I_2 = \pi \int_{-i\infty}^{i\infty} dz e^{a z + b^2 z^2} \left( \csc(\pi z) -\frac{1}{\pi z} \right), $$ where we have set $\epsilon=0$ since the singularity is removed.

Putting $z = i u$ leads to $$ I_2 = 2 i \pi \int_0^{\infty} du\ e^{-b^2 u^2} \sin(a u) \left(\mbox{csch}(\pi u) - \frac{1}{\pi u}\right) .$$ Now expand the exponential, and use the relationship $$ \frac{\partial^{2k}}{\partial a^{2k}} \sin(a u) = (-1)^k u^{2k} sin(a u) $$ to obtain

$$ I_2 = 2 i \pi \sum_{k=0}^{\infty} \frac{b^{2k}}{k!} \frac{\partial^{2k}}{\partial a^{2k}} \int_0^{\infty} du\ \sin(a u) \left( \mbox{csch} (\pi u) - \frac{1}{\pi u} \right). $$

The integral can be evaluated to give $-1/(e^a + 1)$. Replacing it in the series leads to a useful series for small $b$, so that $$ I = i\pi + i\pi\ \mbox{erf}\left( \frac{a}{2 b} \right) - 2 i \pi \sum_{k=0}^{\infty} \frac{b^{2 k}}{k!} \frac{\partial^{2k}}{\partial a^{2k}} \frac{1}{e^a+1} .$$

I (or you) need to analyze the radius of convergence of the series; numerical tests I have done show it to converge quickly for small $b/a$ much less than 1, while for larger values it does not seem to converge. Will get back to you on that.

May I ask, what are you going to use this for?

EDIT 2: The best I have been able to come up with is the following (for small $b$): your integral is equal to, in the limit of zero $\epsilon$,

$$ I = i\pi + 2 i \pi \int_0^{\infty} du\ \frac{\sin(a u)}{\sinh (\pi u)} e^{-b^2 u^2} .$$

The integral can be evaluated for $b=0$: $$ \int_0^{\infty} du\ \frac{\sin(a u)}{\sinh (\pi u)} = \frac{1}{2} \tanh\left( \frac{a}{2} \right) $$.

Now, if $b$ is small, we can expand the exponential and obtain the required integrals by differentiating wrt to $a$; these leads to a series that converges rapidly if $a$ is large enough. That is, the series is useful for small $b/a$. The result is

$$ I \approx i\pi + i\pi \sum_{k=0} \frac{b^{2k}}{k!} \frac{\partial^{2k}}{\partial a^{2k}} \tanh\left( \frac{a}{2} \right) $$

For example, for $a=2$, if we compute the difference between the numerically integrated integral and the sum, taken to 11 terms, for the values $b=1/m, m=1,2,\cdots, 10$, we find

14.568696715793862860, 2.8724585353959*10^-6, 2.410511554*10^-10, 2.718473*10^-13, 1.3485*10^-15, 2.55*10^-17, 2.53*10^-17, 4.93*10^-17, 7.79*10^-17, 1.077*10^-16

So, for $a=2, b=1$ the series is only asymptotic; you get a much better answer using just a couple of terms, but for small $b/a$ it works quite well. I suggest doing some numerical experiments yourself to see if this answer meets your needs.

Hope this helps,

Tom

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Thank you Tom! It will take me some time to digest your solution. This integral is actually the Fourier transform of the following integral: $$ \mathbb{E}[\log(1+e^X)], ~~~ X=\text{Normal}(\mu,\sigma^2) $$ ($\mu,\sigma$ are functions of $a,b$). Unfortunately I'm interested in approximating $F(a,b)$ for small $b$'s. –  user9121 Mar 9 '13 at 0:43
    
OK, knowing that $b$ is small may help, I'll look at it some more. I have fun tackling these types of problems, not always successfully. It seems that this is a tough case because the usual moving of the contour across the poles trick fails. The book "Asymptotics and Mellin-Barnes integrals" - R.B. Paris, D. Kaminski is excellent and treats cases where there is a contour barrier, i. e. where the contour can not be moved past a certain point without giving a divergent integral. Hopefully this can be worked out, Tom –  Tom Dickens Mar 9 '13 at 2:06

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