A complicated integral to be evaluated by complex contour method

[Bengy]

TL;DR Summary

Evaluating a complicated integral

I encounter a very complicated integral in my research project, and it can only be done by complex contour integral. I have tried many other methods like integration by parts and substitution, which all fail. In the integral, all a, alpha and beta are positive real numbers. I have found a similar youtube video which has the same denominator as mine, but the numerator with cosine and exponential is still an issue. I have enclosed the integral as an attachment. Can anyone offer me some hint ? :)

 

[jedishrfu]

For complicated integrals, look for some integration tables like in a CRC math tables book or Schaums outlines math tables book.

They organize integrals by their components like the $\sqrt{x^2-a^2}$ and from there you might a matching integral and solution.

 

[Gavran]

The problem can not be solved analytically.

 

[anuttarasammyak]

for a>0 $\beta \neq 0$ if not mistaken
$$I=\frac{e^{-\frac{\alpha^2}{2\beta}}}{a}\ Re \int_0^\frac{\pi}{2} d\theta \ \ e^{-2\beta a^2(cosec\ \theta-i\frac{\alpha}{2\beta a})^2}$$
For $\alpha=\beta=0, I=\frac{\pi}{2a}$ which is an upper limit of the integral.

[EDIT]
$$I=\frac{1}{8ai}\int_C \frac{dz}{z} \{ \ \exp[ \frac{B}{(z-z^{-1})^2}+\frac{A}{z-z^{-1}}]+ \exp[ \frac{B}{(z-z^{-1})^2}-\frac{A}{z-z^{-1}}] \ \}$$
where closed contour C is the unit circle,
$$A=4a\alpha,\ B=8a^2\beta$$
but I cannot go further.
With the result of @renormalize post #6 for A=0
$$\int_C \frac{dz}{z} \ \exp[ \frac{B}{(z-z^{-1})^2}]=2\pi i \ erfc (\frac{\sqrt{B}}{2})$$
It seems as if $erfc (\frac{\sqrt{B}}{2})$ is the residue at pole z=0. An interesting relation.

 

[Gavran]

For $\alpha=\beta=0, I=\frac{\pi}{2a}$ which is an upper limit of the integral.

This is correct.

The definite integral can be solved analytically by using the substitution method when $\alpha=\beta=0$ and $a\gt0$.

$\begin{align} I&=\int_{a}^{\infty}\frac{\cos(2\alpha x)}{x\sqrt{x^2-a^2}}e^{-2\beta x^2}dx&\nonumber\\ &=\int_{a}^{\infty}\frac{1}{x\sqrt{x^2-a^2}}dx\nonumber \end{align}$

$\begin{align} &t^2=\frac{x^2}{a^2}-1\nonumber\\ & tdt=\frac{x}{a^2}dx\nonumber\\ &|_{a}^{\infty}\to|_{0}^{\infty}\nonumber \end{align}$

$\begin{align} I&=\frac{1}{a}\int_{0}^{\infty}\frac{1}{t^2+1}dt\nonumber\\ &=\frac{1}{a}\arctan t|_{0}^{\infty}\nonumber\\ &=\frac{\pi}{2a}\nonumber \end{align}$

By the way, I still believe that the problem can not be solved analytically when $a$, $\alpha$ and $\beta$ are positive real numbers.

 

[renormalize]

By the way, I still believe that the problem can not be solved analytically when $a$, $\alpha$ and $\beta$ are positive real numbers.

As long as we're posting simpler integrals that can be done, I'd like to nominate:$$\intop_{a}^{\infty}\frac{e^{-2\beta x^{2}}}{x\sqrt{x^{2}-a^{2}}}dx=\frac{\pi}{2a}\text{erfc}\left(\sqrt{2\beta}\,a\right)\quad\left(a>0,\beta>0\right)$$(courtesy of Mathematica)
But I see no way to leverage this result to analytically evaluate the OP's original integral.

 

[Bengy]

Thank you very much for all the replies! Indeed it is a very difficult problem. I find that Mathematica can also give an exact result if I omit the exponential term.

 

[Bengy]

As long as we're posting simpler integrals that can be done, I'd like to nominate:$$\intop_{a}^{\infty}\frac{e^{-2\beta x^{2}}}{x\sqrt{x^{2}-a^{2}}}dx=\frac{\pi}{2a}\text{erfc}\left(\sqrt{2\beta}\,a\right)\quad\left(a>0,\beta>0\right)$$(courtesy of Mathematica)
But I see no way to leverage this result to analytically evaluate the OP's original integral.

Thanks, but the cosine term is important for my case, as my problem has oscillatory behaviour

 

[julian]

I'm not sure if this will be useful for your purposes, but here’s an expansion in $\alpha$:

\begin{align*}
I & = \int_a^\infty \dfrac{\cos (2 \alpha x)}{x \sqrt{x^2-a^2}} e^{- 2 \beta x^2} dx
\nonumber \\
& = \int_a^\infty \dfrac{\sum_{n=0}^\infty \frac{1}{{(2n)}!} (-1)^n (2 \alpha x)^{2n}}{x \sqrt{x^2-a^2}} e^{- 2 \beta x^2} dx
\nonumber \\
& = \sum_{n=0}^\infty \frac{1}{{(2n)}!} (-1)^n (2 \alpha)^{2n} \int_a^\infty \dfrac{x^{2n}}{x \sqrt{x^2-a^2}} e^{- 2 \beta x^2} dx
\nonumber \\
& = \sum_{n=0}^\infty \frac{1}{{(2n)}!} (-1)^n (2 \alpha)^{2n} \cdot \frac{(-1)^n}{2^n} \frac{\partial^n}{\partial \beta^n} \int_a^\infty \dfrac{e^{- 2 \beta x^2}}{x \sqrt{x^2-a^2}} dx
\nonumber \\
& = \frac{\pi}{2a} \sum_{n=0}^\infty \frac{1}{{(2n)}!} \left( 2 \alpha^2 \right)^n \frac{\partial^n}{\partial \beta^n} \text{erfc} (\sqrt{2 \beta} a)
\nonumber \\
& = \frac{\pi}{2a} \sum_{n=0}^\infty \frac{1}{{(2n)}!} \left( 2 \alpha^2 \right)^n \frac{2}{\sqrt{\pi}} \frac{\partial^n}{\partial \beta^n} \int_{\sqrt{2 \beta} a}^\infty e^{- t^2} dt
\nonumber \\
& = \frac{\pi}{2a} \sum_{n=0}^\infty \frac{1}{{(2n)}!} \left( 2 \alpha^2 \right)^n \frac{2}{\sqrt{\pi}} \frac{\partial^n}{\partial \beta^n} \frac{1}{2} \int_{2 \beta a^2}^\infty e^{- u} \frac{du}{\sqrt{u}}
\nonumber \\
& = \frac{\pi}{2a} \text{erfc} (\sqrt{2 \beta} a) - \sqrt{\pi} a \sum_{n=1}^\infty \frac{2^n}{{(2n)}!} \alpha^{2n} \frac{\partial^{n-1}}{\partial \beta^{n-1}} \frac{e^{- 2 \beta a^2}}{\sqrt{2 \beta a^2}}
\end{align*}

For those interested, here’s a method to prove the integral: $I=\int_a^\infty \dfrac{e^{- 2 \beta x^2}}{x \sqrt{x^2-a^2}} dx = \frac{\pi}{2a} \text{erfc} (\sqrt{2 \beta} a)$ :

Use $u^2=x^2-a^2$, then

\begin{align*}
I & = e^{- 2 \beta a^2} \int_0^\infty \dfrac{e^{- 2 \beta u^2}}{u \sqrt{u^2+a^2}} \frac{udu}{\sqrt{u^2+a^2}}
\nonumber \\
& = e^{- 2 \beta a^2} \int_0^\infty \dfrac{e^{- 2 \beta u^2}}{u^2+a^2} du
\end{align*}

Then use $\frac{1}{u^2+a^2}= \int_0^\infty e^{- (u^2+a^2) s} ds$:

\begin{align*}
I & = e^{- 2 \beta a^2} \int_0^\infty e^{- 2 \beta u^2} \int_0^\infty e^{- (u^2+a^2) s} ds du
\nonumber \\
& = e^{- 2 \beta a^2} \int_0^\infty e^{- a^2 s} \int_0^\infty e^{- u^2 (s+ 2 \beta)} du ds
\nonumber \\
& = e^{- 2 \beta a^2} \int_0^\infty e^{- a^2 s} \int_0^\infty e^{- u^2} \frac{du}{\sqrt{s+ 2 \beta}} ds
\nonumber \\
& = e^{- 2 \beta a^2} \frac{\sqrt{\pi}}{2} \int_0^\infty \frac{e^{- a^2 s}}{\sqrt{s+ 2 \beta}} ds
\end{align*}

Use $t=\sqrt{s+ 2 \beta}$, then $dt = \frac{ds}{2 \sqrt{s+ 2 \beta}}$ and:

\begin{align*}
I & = e^{- 2 \beta a^2} \sqrt{\pi} \int_{\sqrt{2 \beta}}^\infty e^{- a^2 (t^2 - 2 \beta)} dt
\nonumber \\
& = \sqrt{\pi} \int_{\sqrt{2 \beta}}^\infty e^{- a^2 t^2} dt
\nonumber \\
& = \frac{\sqrt{\pi}}{a} \int_{\sqrt{2 \beta} a}^\infty e^{- t^2} dt
\nonumber \\
& = \frac{\pi}{2a} \text{erfc} (\sqrt{2 \beta} a) .
\end{align*}

 

[Bengy]

Thank you very much for your effort! It provides more insight to the question.

 

[julian]

for a>0 $\beta \neq 0$ if not mistaken
$$I=\frac{e^{-\frac{\alpha^2}{2\beta}}}{a}\ Re \int_0^\frac{\pi}{2} d\theta \ \ e^{-2\beta a^2(cosec\ \theta-i\frac{\alpha}{2\beta a})^2}$$
For $\alpha=\beta=0, I=\frac{\pi}{2a}$ which is an upper limit of the integral.

[EDIT]
$$I=\frac{1}{8ai}\int_C \frac{dz}{z} \{ \ \exp[ \frac{B}{(z-z^{-1})^2}+\frac{A}{z-z^{-1}}]+ \exp[ \frac{B}{(z-z^{-1})^2}-\frac{A}{z-z^{-1}}] \ \}$$
where closed contour C is the unit circle,
$$A=4a\alpha,\ B=8a^2\beta$$
but I cannot go further.
With the result of @renormalize post #6 for A=0
$$\int_C \frac{dz}{z} \ \exp[ \frac{B}{(z-z^{-1})^2}]=2\pi i \ erfc (\frac{\sqrt{B}}{2})$$
It seems as if $erfc (\frac{\sqrt{B}}{2})$ is the residue at pole z=0. An interesting relation.

The function $\exp [\frac{B}{(z-z^{-1})^2}] = \exp [\frac{B z^2}{(1-z^2)^2}]$ doesn't seem to be singular at $z=0$. The poles appear to be on the contour $C$ at $z = \pm 1$ instead.

Additionally, the result of expanding $\exp [\frac{B z^2}{(1-z^2)^2}]$ and dealing with each term at a time: $\frac{1}{4ai} \int_C \frac{dz}{z} \exp [\frac{B z^2}{(1-z^2)^2}] = \frac{1}{4ai} \sum_{n=0}^\infty \frac{B^n}{n!} \oint_C \frac{dz}{z} \frac{z^{2n}}{(1-z^2)^{2n}}$ couldn't match the Taylor series you would be aiming for, which is:

\begin{align*}
\frac{\pi}{2a} \text{erfc} (\sqrt{B}/2) & = \frac{\pi}{2a} (1-\text{erf} (\sqrt{B}/2) ) = \frac{\pi}{2a} (1- \frac{2}{\sqrt{\pi}} \int_0^{\sqrt{B}/2} e^{-t^2} dt)
\nonumber \\
& = \frac{\pi}{2a} (1-\frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty \frac{(-1)^n}{n!} \int_0^{\sqrt{B}/2} t^{2n} dt)
\nonumber \\
& = \frac{\pi}{2a} (1-\frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty \frac{(-1)^n}{n!} \left[ \frac{t^{2n+1}}{2n+1} \right]_0^{\sqrt{B}/2})
\nonumber \\
& = \frac{\pi}{2a} - \frac{\sqrt{\pi}}{2a} \sqrt{B} \sum_{n=0}^\infty \frac{(-1)^n}{2^n n! (2n+1)} B^n
\end{align*}

because of the factor of $\sqrt{B}$ outside the sum. This means you can't evaluate your complex contour integral by first expanding the exponential. It would be interesting to see how to deal with these issues.

Expanding the exponetial your $\theta$ integral doesn't work as you get divergent integrals:

I agree with your $\theta$ integral. However, putting $\alpha=0$ in your integral representation, and then expanding the exponetial would give:

\begin{align*}
I & = \frac{1}{a} Re \int_0^{\pi/2} e^{-2 \beta a^2 cosec^2 \theta} d \theta
\nonumber \\
& = \frac{1}{a} \sum_{n=0}^\infty \frac{(-1)^n (2 \beta a^2)^n}{n!} \int_0^{\pi/2} cosec^{2n} \theta d \theta
\nonumber \\
& = \frac{\pi}{2a} + \frac{1}{a} \sum_{n=0}^\infty \frac{(-1)^{n+1} (2 \beta a^2)^{n+1}}{(n+1)!} \int_0^{\pi/2} cosec^{2n+2} \theta d \theta
\nonumber \\
& = \frac{\pi}{2a} + 2 a \beta \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(n+1)!} (2 \beta a^2)^n \int_0^{\pi/2} cosec^{2n+2} \theta d \theta
\end{align*}

but the Talyor series expansion for $\frac{\pi}{2a} \text{erfc} (\sqrt{2 \beta} a)$ is $\frac{\pi}{2a} - \sqrt{2 \pi}\sqrt{\beta} \sum_{n=0}^\infty \frac{(-1)^n}{n! (2n+1)} (2 \beta a^2)^n$. We get the same issue as before, that there is a $\sqrt{\beta}$ outside the sum.

The reason this approach can't work is because all the cosec integrals diverge: First, $\frac{d}{d \theta} \cot \theta = - cosec^2 \theta$ implies

\begin{align*}
\int_0^{\pi/2} cosec^2 \theta d \theta = \left[ - \cot \theta \right]_0^{\pi/2} = \infty
\end{align*}

Then as $cosec^2 \theta < cosec^{2n+2} \theta$ implies that $\infty = \int_0^{\pi/2} cosec^2 \theta d \theta < \int_0^{\pi/2} cosec^{2n+2} \theta d \theta$.

Noting

\begin{align*}
\infty = \frac{1}{2^{2n}} \int_0^{2\pi} cosec^{2n} \theta d \theta = \oint_C \frac{dz}{z} \frac{z^{2n}}{(1-z^2)^{2n}}
\end{align*}

and ignoring the issue of poles on the contour, the contour integral diverges, which is why expanding the exponential $\frac{1}{4ai} \int_C \frac{dz}{z} \exp [\frac{B z^2}{(1-z^2)^2}]$doesn't work.

I agree with your $\theta$ integral:

A minor point: I get the same answer as you did for their integral expression: Using $x=a cosec \theta$

\begin{align*}
I & = e^{-\alpha^2/2\beta} Re \int_a^\infty \frac{e^{-2 \beta (x - i \alpha/2\beta)^2}}{x \sqrt{x^2-a^2}} dx
\nonumber \\
& = e^{-\alpha^2/2\beta} Re \int_{\pi/2}^0 \frac{e^{-2 \beta (a cosec \theta - i \alpha/2\beta)^2}}{a^2 cosec \theta \cot \theta} \cdot - a \cot \theta cosec \theta d \theta
\nonumber \\
& = \frac{e^{-\alpha^2/2\beta}}{a} Re \int_0^{\pi/2} e^{-2 \beta a^2 (cosec \theta - i \alpha/2a\beta)^2} d \theta
\end{align*}

I'm not quite getting the same result as you for the next part:

\begin{align*}
I & = \frac{e^{-\alpha^2/2\beta}}{2a} (\int_0^{\pi/2} e^{-2 \beta a^2 (cosec \theta - i \alpha/2a\beta)^2} d \theta + \int_0^{\pi/2} e^{-2 \beta a^2 (cosec \theta + i \alpha/2a\beta)^2} d \theta)
\nonumber \\
& = \frac{e^{-\alpha^2/2\beta}}{4a} (\int_0^\pi e^{-2 \beta a^2 (cosec \theta - i \alpha/2a\beta)^2} d \theta + \int_\pi^{2\pi} e^{-2 \beta a^2 (cosec \theta - i \alpha/2a\beta)^2} d \theta)
\nonumber \\
& = \frac{e^{-\alpha^2/2\beta}}{4a} \int_0^{2\pi} e^{-2 \beta a^2 (cosec \theta - i \alpha/2a\beta)^2} d \theta
\nonumber \\
& = \frac{e^{-\alpha^2/2\beta}}{4a i} \oint_C e^{- 2 \beta a^2 (2i(z-z^{-1})^{-1} - i \alpha/2a\beta)^2} \frac{d z}{z}
\nonumber \\
& = \frac{1}{4a i} \oint_C e^{8\beta a^2 [(z-z^{-1})^2 + (z-z^{-1}) \alpha/2a\beta)} \frac{d z}{z}
\nonumber \\
& = \frac{1}{4a i} \oint_C \frac{d z}{z} \exp [8\beta a^2 (z-z^{-1})^{-2} + 4 a \alpha (z-z^{-1})^{-1}]
\nonumber \\
& = \frac{1}{4a i} \oint_C \frac{d z}{z} \exp [\frac{B}{(z-z^{-1})^2} + \frac{A}{(z-z^{-1})}]
\end{align*}

where I have used $cosec \theta$ is symmetric around $\pi/2$, $cosec (\theta+\pi)=-cosec \theta$, and put aside subtlies about poles being on the contour.

 

[anuttarasammyak]

@julian Thank you so much for fully detailed consideration.
What I had thought. $$z-z^{-1}$$ is pure imaginary for z on C, so exp of B does not diverge but goes to zero at $z=\pm 1$. So they do not seem to be singularities. But I do not know how to use it.

 

[julian]

But if you put

\begin{align*}
e^{i \theta} - e^{-i \theta} = 2 i \sin \theta = 0
\end{align*}

$\theta=0$ and $\theta=\pi$ are clearly solutions.

 

[anuttarasammyak]

I would say
$$exp[\frac{B}{(2i \sin\theta)^2}]=exp[-\frac{B}{4 \sin^2\theta}]$$
It goes to zero when approaching to $\theta=0,\pi$.

 

[julian]

Oh, I see. Accumulated fatigue from the past few days.