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I have a really nasty integral to solve as follow: (It is good for challenge lovers!)

$$ I(t)=\int_{0}^{t}{{\bigg(\cos\left({\frac{\gamma}{4(t-r)}}\right)+\sin\left({\frac{\gamma}{4(t-r)}}\right)\bigg)} e^{- \frac{\alpha}{4r}} \left(\frac{\sqrt{\alpha}-2\sqrt{2} \lambda r}{4\pi r \sqrt{r(t-r)}} \right) }~\mathrm{d}r. \tag1 $$

I applied two change of variables as $x=\frac{\gamma}{4(t-r)}$ and $u^2=x-\frac{\gamma}{4t}$, respectively. Finally, I obtain

$$ I(t)= \int_{0}^{\infty}{ \left( \cos \left( {\frac{\gamma}{4t}+u^2} \right)+\sin \left( {\frac{\gamma}{4t}+u^2} \right) \right) e^{-\frac{C}{u^2}} \left(\frac{A}{4tu^2+\gamma} + \frac{B}{u^2}\right) }~\mathrm{d}u \tag2 $$

where $A$, $B$, $C$, $\gamma, \alpha, \lambda$ and $t$ are constants.

How can I solve the last integral ?

Thanks!

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  • $\begingroup$ Are you sure the integral converges? $\endgroup$ – Ron Gordon Aug 11 '15 at 13:24
  • $\begingroup$ @RonGordon : I checked my calculations and I edited again. I think it is convergent now. $\endgroup$ – Wita Aug 12 '15 at 15:32
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I'm looking at integral (1) and seeing a Laplace transform convolution integral. Note that

$$\int_0^t dt' \, f(t') g(t-t') = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, F(s) G(s) e^{s t}$$

where

$$F(s) = \int_0^{\infty} dt \, f(t) e^{-s t} $$ $$G(s) = \int_0^{\infty} dt \, g(t) e^{-s t} $$

In this case, we may define

$$f_1(t) = t^{-3/2} e^{-\frac{a}{4 t}} $$ $$f_2(t) = t^{-1/2} e^{-\frac{a}{4 t}} $$ $$g_1(t) = t^{-1/2} \cos{\left ( \frac{b}{2 t} \right )} $$ $$g_2(t) = t^{-1/2} \sin{\left ( \frac{b}{2 t} \right )} $$

Then (I will show this if necessary)

$$F_1(s) = 2 \sqrt{\frac{\pi}{a}} e^{-\sqrt{a s}}$$ $$F_2(s) = \sqrt{\pi} s^{-1/2} e^{-\sqrt{a s}}$$ $$G_1(s) = \sqrt{\pi} s^{-1/2} e^{-\sqrt{b s}} \cos{\sqrt{b s}}$$ $$G_2(s) = \sqrt{\pi} s^{-1/2} e^{-\sqrt{b s}} \sin{\sqrt{b s}}$$

So let's evaluate

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-1/2} e^{-\sqrt{p s}} \cos{\sqrt{b s}} \, e^{s t}$$

where $\sqrt{p} = \sqrt{a}+\sqrt{b} $. To evaluate this, we use Cauchy's theorem and consider the contour integral

$$\oint_C dz \, z^{-1/2} e^{-\sqrt{p z}} \cos{\sqrt{b z}} \, e^{z t}$$

where $C$ is the following contour:

enter image description here

We will define $\text{Arg}{z} \in (-\pi,\pi]$, so the branch is the negative real axis. There are $6$ pieces to this contour, $C_k$, $k \in \{1,2,3,4,5,6\}$, as follows.

$C_1$ is the contour along the line $z \in [c-i R,c+i R]$ for some large value of $R$.

$C_2$ is the contour along a circular arc of radius $R$ from the top of $C_1$ to just above the negative real axis.

$C_3$ is the contour along a line just above the negative real axis between $[-R, -\epsilon]$ for some small $\epsilon$.

$C_4$ is the contour along a circular arc of radius $\epsilon$ about the origin.

$C_5$ is the contour along a line just below the negative real axis between $[-\epsilon,-R]$.

$C_6$ is the contour along the circular arc of radius $R$ from just below the negative real axis to the bottom of $C_1$.

We will show that the integral along $C_2$,$C_4$, and $C_6$ vanish in the limits of $R \rightarrow \infty$ and $\epsilon \rightarrow 0$.

On $C_2$, the real part of the argument of the exponential is

$$R t \cos{\theta} - \sqrt{p R} \cos{\frac{\theta}{2}} + \sqrt{b R} \sin{\frac{\theta}{2}}$$

where $\theta \in [\pi/2,\pi)$. Clearly, $\cos{\theta} < 0$, $\cos{\frac{\theta}{2}} > 0$, and $\sin{\frac{\theta}{2}} > 0$ so that the integrand exponentially decays as $R \rightarrow \infty$ and therefore the integral vanishes along $C_2$.

On $C_6$, we have the same thing, but now $\theta \in (-\pi,-\pi/2]$. This means that, due to the evenness of cosine, the integrand exponentially decays again as $R \rightarrow \infty$ and therefore the integral also vanishes along $C_6$.

On $C_4$, the integral vanishes as $\epsilon$ in the limit $\epsilon \rightarrow 0$. Thus, we are left with the following by Cauchy's integral theorem (i.e., no poles inside $C$):

$$\left [ \int_{C_1} + \int_{C_3} + \int_{C_5}\right] dz \: z^{-1/2} e^{-\sqrt{p z}} \cos{\sqrt{b z}} \, e^{z t} = 0$$

On $C_3$, we parametrize by $z=e^{i \pi} x$ and the integral along $C_3$ becomes

$$\int_{C_3} dz \: z^{-1/2} e^{-\sqrt{p z}} \cos{\sqrt{b z}} \, e^{z t} = e^{i \pi} \int_{\infty}^0 dx \: e^{-i \pi/2} x^{-1/2} e^{-i \sqrt{p x}} \cosh{\sqrt{b x}} \, e^{-x t}$$

On $C_5$, however, we parametrize by $z=e^{-i \pi} x$ and the integral along $C_5$ becomes

$$\int_{C_5} dz \: z^{-1/2} e^{-\sqrt{p z}} \cos{\sqrt{b z}} \, e^{z t} = e^{-i \pi} \int_0^{\infty} dx \: e^{i \pi/2} x^{-1/2} e^{i \sqrt{p x}} \cosh{\sqrt{b x}} \, e^{-x t}$$

We may now write, by Cauchy's theorem,

$$-\frac{1}{2 \pi} \int_0^{\infty} dx \: x^{-1/2} \, e^{- x t} \left ( e^{i \sqrt{p x}} + e^{-i \sqrt{p x}} \right ) \cosh{\sqrt{b x}} + \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1/2} \, e^{-\sqrt{p s}} \cos{\sqrt{b s}} \, e^{s t} = 0$$

Therefore, the ILT of $\hat{f}(s) = s^{-1/2} \, e^{-\sqrt{p s}} \cos{\sqrt{b s}}$ is given by

$$\begin{align}\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: e^{-\sqrt{s}} e^{s t} &= \frac{1}{\pi} \int_0^{\infty} dx \: x^{-1/2} \,e^{- x t} \, \cos{\sqrt{p x}} \, \cosh{\sqrt{b x}}\\ &= \frac{1}{\pi} \int_{-\infty}^{\infty} du\: e^{-t u^2} \cosh{\sqrt{b} u} \cos{\sqrt{p} u}\end{align}$$

The last step involved substituting $x=u^2$ and exploiting the evenness of the integrand.

Therefore the result is that

$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1/2} \, e^{-\sqrt{p s}}\, \cos{\sqrt{b s}} \, e^{s t} = (\pi t)^{-1/2} e^{\frac{b-p}{4 t}} \cos{\left ( \frac{\sqrt{b p}}{2 t} \right )}$$

Similarly,

$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1/2} \, e^{-\sqrt{p s}}\, \sin{\sqrt{b s}} \, e^{s t} = (\pi t)^{-1/2} e^{\frac{b-p}{4 t}} \sin{\left ( \frac{\sqrt{b p}}{2 t} \right )}$$

We also need to find the ILTs

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1} \, e^{-\sqrt{p s}}\, \cos{\sqrt{b s}} \, e^{s t} $$

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1} \, e^{-\sqrt{p s}}\, \sin{\sqrt{b s}} \, e^{s t} $$

For similar reasons as above, we may write

$$\frac{1}{i 2 \pi} \int_0^{\infty} dx \: x^{-1} \, e^{- x t} \left ( e^{i \sqrt{p x}} - e^{-i \sqrt{p x}} \right ) \cosh{\sqrt{b x}} + \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1} \, e^{-\sqrt{p s}} \cos{\sqrt{b s}} \, e^{s t} = 1$$

$$-\frac{1}{2 \pi} \int_0^{\infty} dx \: x^{-1} \, e^{- x t} \left ( e^{i \sqrt{p x}} + e^{-i \sqrt{p x}} \right ) \sinh{\sqrt{b x}} + \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1} \, e^{-\sqrt{p s}} \sin { \sqrt{b s} } \, e^{s t} = 0$$

Note that, in the former equation, the integral around $C_4$ is nonzero.

Thus,

$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1} \, e^{-\sqrt{p s}} \cos{\sqrt{b s}} \, e^{s t} = 1 - \frac1{\pi} \int_{-\infty}^{\infty} du\: e^{-t u^2} \cosh{\sqrt{b} u} \frac{\sin{\sqrt{p} u}}{u} $$

$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1} \, e^{-\sqrt{p s}} \sin{\sqrt{b s}} \, e^{s t} = \frac1{\pi} \int_{-\infty}^{\infty} du\: e^{-t u^2} \cos{\sqrt{p} u} \frac{\sinh{\sqrt{b} u}}{u} $$

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  • $\begingroup$ Thank you for your answer. However, I would be pleased if you can give more detail about f1, f2, g1, and g2. Because you used $a$ and $b$ as a variable and I do not see $\lambda$ in your calculations. $\endgroup$ – Wita Aug 13 '15 at 15:52
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    $\begingroup$ @2che: yes, sorry, this is sort of a work in progress. I just wanted to outline what I am doing. I will finish and fill in the blanks. That said, it shouldn't be too hard to see that $a=\alpha$ and $b=\gamma/2$. I am taming the notation so the Laplace transform work is less cluttered. $\endgroup$ – Ron Gordon Aug 13 '15 at 16:17
  • $\begingroup$ Can we write last 2 integrals with a simpler form ? $\endgroup$ – Wita Oct 12 '15 at 15:42
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For $C>0$ and real, Mathematica gives the answer to the $I(t)$ integral as:

$\frac{\sqrt{\pi } e^{-\sqrt{2} \sqrt{C}} \left(\left(2 A C+2 \sqrt{2} B \sqrt{C}+B\right) \sin \left(\sqrt{2} \sqrt{C}+\frac{\gamma }{4 t}\right)+(2 A C+B) \cos \left(\sqrt{2} \sqrt{C}+\frac{\gamma }{4 t}\right)\right)}{16 C^{3/2} t}$

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