- #541
saltydog
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I wish to finally evaluate:
<br /> y(t)=\frac{e^{-\pi/t}}{2 i t^{3/2}}\mathop\lim\limits_{R\to\infty}\left\{\text{Erfi}\left(v-\sqrt{\pi/t}\right)_{\sqrt{Rt}e^{-\pi i/4}}^{\sqrt{Rt}e^{\pi i/4}}\right\}\tag{1}<br />
First note:
<br /> \text{Erfi}(z)=\frac{\text{Erf(iz)}}{i}=\frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{iz} e^{-w^2}dw<br />
Using this relation, I can re-write the above limit as:
<br /> \frac{2}{i\sqrt{\pi}}\left\{\mathop\lim\limits_{R\to\infty}\int_0^{i(\sqrt{Rt}e^{\pi i/4})} e^{-w^2}dw-\mathop\lim\limits_{R\to\infty}\int_0^{i(\sqrt{Rt}e^{-\pi i/4})} e^{-w^2}dw\right\}<br />
Using Euler's relation, I convert the upper limits to real and imaginary parts and multiply by i:
<br /> \begin{align*}<br /> i\sqrt{Rt}e^{\pi i/4}&=i\left[\left(\sqrt{Rt/2}-\sqrt{\pi/2}\right)+i\sqrt{Rt/2}\right] \\<br /> &=-\sqrt{Rt/2}+i\left(\sqrt{Rt/2}-\sqrt{\pi/t}\right)<br /> \end{align}<br />
Now, as R goes to infinity, this quantity goes to -\infty+i\infty.
And the lower limit:
<br /> \begin{align*}<br /> i\sqrt{Rt}e^{-\pi i/4}&=i\left[\sqrt{Rt/2}+\left(\sqrt{Rt/2}-\sqrt{\pi/2}\right)\right] \\<br /> &=\sqrt{Rt/2}+i\left(\sqrt{Rt/2}-\sqrt{\pi/2}\right)<br /> \end{align}<br />
as R goes to infinity, this limit goes to \infty+i\infty.
Thus the integrals can be rewritten as:
<br /> \frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{-\infty+i\infty} e^{-w^2}dw-<br /> \frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{\infty+i\infty} e^{-w^2}dw<br />
I choose to evaluate these line integrals along a parametric path using the following relation from Complex Analysis:
<br /> \int_C f(z)dz=\int_c udx-vdy+i\int_c vdx+udy<br />
Thus the first step is to express the integrand in terms of a real and imaginary part. Relying on Euler's formula:
<br /> \begin{align*}<br /> e^{-w^2}&=e^{-(x+yi)^2} \\<br /> &=e^{-(x^2-y^2+2xyi)} \\<br /> &=e^{-(x^2-y^2)}Cos(2xy)+ie^{-(x^2-y^2)}(-Sin(2xy))\\<br /> &=u+iv<br /> \end{align}<br />
For the first integral, the path goes from the origin to the point -\infty+i\infty
Thus if I let:
x=-t,\quad dx=-dt
y=t,\quad dy=dt
Making this substitution, note that the exponent term is 1 and we have for the first integral (with t going from 0 to infinity):
<br /> \begin{align*}<br /> \int_0^{-\infty+i\infty} e^{-w^2}dw&=\left(-\int_0^\infty Cos(-2t^2)dt+\int_0^\infty Sin(-2t^2)dt\right) \\<br /> &+i\left(\int_0^\infty Cos(-2t^2)dt+\int_0^\infty Sin(-2t^2)dt\right) \\<br /> &=\left(-\frac{\sqrt{\pi}}{4}-\frac{\sqrt{\pi}}{4}\right)+i\left(\frac{\sqrt{\pi}}{4}-\frac{\sqrt{\pi}}{4}\right) \\<br /> &=-\frac{\sqrt{\pi}}{2}<br /> \end{align}<br />
The same analysis can be made (with the path going from 0 to \infty+i\infty for the second integral with the results that the line integral in that case is:
<br /> \frac{\sqrt{\pi}}{2}
Substituting these values into (1), I obtain:
<br /> \frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{-\infty+i\infty} e^{-w^2}dw=\frac{1}{i}\frac{2}{\sqrt{\pi}}(-\frac{\sqrt{\pi}}{2})=-\frac{1}{i}\frac{i}{i}=i<br />
and:
<br /> \frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{\infty+i\infty} e^{-w^2}dw=\frac{1}{i}\frac{2}{\sqrt{\pi}}(\frac{\sqrt{\pi}}{2})=\frac{1}{i}\frac{i}{i}=-i<br />
So that the limit is 2i.
Substituting this into the last expression of the inverse transform, we obtain:
<br /> y(x)=\frac{e^{-\pi/t}}{2i t^{3/2}}(2i)=\frac{e^{-\pi/t}}{t^{3/2}}
<br /> y(t)=\frac{e^{-\pi/t}}{2 i t^{3/2}}\mathop\lim\limits_{R\to\infty}\left\{\text{Erfi}\left(v-\sqrt{\pi/t}\right)_{\sqrt{Rt}e^{-\pi i/4}}^{\sqrt{Rt}e^{\pi i/4}}\right\}\tag{1}<br />
First note:
<br /> \text{Erfi}(z)=\frac{\text{Erf(iz)}}{i}=\frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{iz} e^{-w^2}dw<br />
Using this relation, I can re-write the above limit as:
<br /> \frac{2}{i\sqrt{\pi}}\left\{\mathop\lim\limits_{R\to\infty}\int_0^{i(\sqrt{Rt}e^{\pi i/4})} e^{-w^2}dw-\mathop\lim\limits_{R\to\infty}\int_0^{i(\sqrt{Rt}e^{-\pi i/4})} e^{-w^2}dw\right\}<br />
Using Euler's relation, I convert the upper limits to real and imaginary parts and multiply by i:
<br /> \begin{align*}<br /> i\sqrt{Rt}e^{\pi i/4}&=i\left[\left(\sqrt{Rt/2}-\sqrt{\pi/2}\right)+i\sqrt{Rt/2}\right] \\<br /> &=-\sqrt{Rt/2}+i\left(\sqrt{Rt/2}-\sqrt{\pi/t}\right)<br /> \end{align}<br />
Now, as R goes to infinity, this quantity goes to -\infty+i\infty.
And the lower limit:
<br /> \begin{align*}<br /> i\sqrt{Rt}e^{-\pi i/4}&=i\left[\sqrt{Rt/2}+\left(\sqrt{Rt/2}-\sqrt{\pi/2}\right)\right] \\<br /> &=\sqrt{Rt/2}+i\left(\sqrt{Rt/2}-\sqrt{\pi/2}\right)<br /> \end{align}<br />
as R goes to infinity, this limit goes to \infty+i\infty.
Thus the integrals can be rewritten as:
<br /> \frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{-\infty+i\infty} e^{-w^2}dw-<br /> \frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{\infty+i\infty} e^{-w^2}dw<br />
I choose to evaluate these line integrals along a parametric path using the following relation from Complex Analysis:
<br /> \int_C f(z)dz=\int_c udx-vdy+i\int_c vdx+udy<br />
Thus the first step is to express the integrand in terms of a real and imaginary part. Relying on Euler's formula:
<br /> \begin{align*}<br /> e^{-w^2}&=e^{-(x+yi)^2} \\<br /> &=e^{-(x^2-y^2+2xyi)} \\<br /> &=e^{-(x^2-y^2)}Cos(2xy)+ie^{-(x^2-y^2)}(-Sin(2xy))\\<br /> &=u+iv<br /> \end{align}<br />
For the first integral, the path goes from the origin to the point -\infty+i\infty
Thus if I let:
x=-t,\quad dx=-dt
y=t,\quad dy=dt
Making this substitution, note that the exponent term is 1 and we have for the first integral (with t going from 0 to infinity):
<br /> \begin{align*}<br /> \int_0^{-\infty+i\infty} e^{-w^2}dw&=\left(-\int_0^\infty Cos(-2t^2)dt+\int_0^\infty Sin(-2t^2)dt\right) \\<br /> &+i\left(\int_0^\infty Cos(-2t^2)dt+\int_0^\infty Sin(-2t^2)dt\right) \\<br /> &=\left(-\frac{\sqrt{\pi}}{4}-\frac{\sqrt{\pi}}{4}\right)+i\left(\frac{\sqrt{\pi}}{4}-\frac{\sqrt{\pi}}{4}\right) \\<br /> &=-\frac{\sqrt{\pi}}{2}<br /> \end{align}<br />
The same analysis can be made (with the path going from 0 to \infty+i\infty for the second integral with the results that the line integral in that case is:
<br /> \frac{\sqrt{\pi}}{2}
Substituting these values into (1), I obtain:
<br /> \frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{-\infty+i\infty} e^{-w^2}dw=\frac{1}{i}\frac{2}{\sqrt{\pi}}(-\frac{\sqrt{\pi}}{2})=-\frac{1}{i}\frac{i}{i}=i<br />
and:
<br /> \frac{1}{i}\frac{2}{\sqrt{\pi}}\int_0^{\infty+i\infty} e^{-w^2}dw=\frac{1}{i}\frac{2}{\sqrt{\pi}}(\frac{\sqrt{\pi}}{2})=\frac{1}{i}\frac{i}{i}=-i<br />
So that the limit is 2i.
Substituting this into the last expression of the inverse transform, we obtain:
<br /> y(x)=\frac{e^{-\pi/t}}{2i t^{3/2}}(2i)=\frac{e^{-\pi/t}}{t^{3/2}}
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