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I am trying to solve

\begin{equation} u(x,t)=\frac{6\sqrt{2}}{\sqrt{\pi}}\int_0^\infty \frac{\sin3\lambda}{\lambda}\cos(\lambda x)e^{-\frac{\lambda^2}{4}t}d\lambda \end{equation}

which gives on Wolfram Alpha:

\begin{equation} u(x,t)=3\sqrt{\frac{{\pi}}{2}}\Bigg[\operatorname{erf}\bigg(\frac{3-x}{\sqrt{t}}\bigg)+\operatorname{erf}\bigg(\frac{3+x}{\sqrt{t}}\bigg)\Bigg] \end{equation}

I try to re-write this into complex terms, where $\lambda$ is the variable and $x$ and 3 are mere coefficients. The first two terms (assuming this is correct extrapolation of the Eulers formula for sine and cosine):

\begin{equation} \frac{\sin3\lambda}{\lambda}\cos(\lambda x)=\frac{e^{3i\lambda}-e^{-3i\lambda}}{2i\lambda}\frac{e^{ix\lambda}+e^{-ix\lambda}}{2} \end{equation}

Thus I get

\begin{equation} u(x,t)=\frac{6\sqrt{2}}{\sqrt{\pi}}\int_0^\infty \frac{e^{3i\lambda}-e^{-3i\lambda}}{2i\lambda}\frac{e^{ix\lambda}+e^{-ix\lambda}}{2} e^{-\frac{\lambda^2}{4}t}d\lambda \end{equation}

Then I set by substitution $u=\lambda^2\rightarrow \lambda=\sqrt{u}$, and $du=2\lambda d\lambda\rightarrow d\lambda=\frac{du}{2\lambda}$

It generates a simpler form:

\begin{equation} \frac{6\sqrt{2}}{\sqrt{\pi}i}\bigg[\int_0^\infty e^{a\sqrt{u}}e^{-\frac{u}{4}t}du+\int_0^\infty e^{b\sqrt{u}}e^{-\frac{u}{4}t}du-\int_0^\infty e^{c\sqrt{u}}e^{-\frac{u}{4}t}du-\int_0^\infty e^{d\sqrt{u}}e^{-\frac{u}{4}t}du\bigg] \end{equation}

where, $a, b, c, d$ are the coefficients in the exponents $a=3i+ix, \ b=3i-ix,\ c=ix-3i,\ d=-3i-ix$

But I can't get further

Any ideas?

Thanks

Gary
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Luthier415Hz
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    What you are supposed to obtain is $$u(x,t)=3 \sqrt{\frac{\pi }{2}} \Bigg[\text{erf}\left(\frac{3-x}{\sqrt{t}}\right)+\text{erf}\left(\frac{3+x}{\sqrt {t}}\right)\Bigg]$$ – Claude Leibovici May 04 '22 at 12:36
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    Write $\cos(\lambda x)$ instead of $cos \lambda x$. Use \operatorname{erf} to produce the correct erf. – Gary May 04 '22 at 12:54
  • Then write $u(\lambda,t)$ – Claude Leibovici May 04 '22 at 12:56
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    The result of the integral cannot depend on $\lambda$ because $\lambda$ is the (dummy) integration variable. It is a function of $x$, Claude is right. – Gary May 04 '22 at 12:59
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    But if you (definite) integrate with respect to $\lambda$ the result is not a function of $\lambda$. This is high school maths. Anyway, I fixed it for you. – Gary May 04 '22 at 13:02
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    That is plain wrong to do. You would just constantly change the role of $\lambda$, making a huge confusion. Believe me, I am doing integration for 15+ years now. – Gary May 04 '22 at 13:03
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    Ok, if you are right then substitute $\lambda=3$ on both sides in your first formula and show me what it looks like. – Gary May 04 '22 at 13:07
  • I don't think I know what you mean. Please make the editions you deem necessary. – Luthier415Hz May 04 '22 at 13:10
  • If you take $\lambda=3$ in your first formula you get $$u(3,t)=\frac{6\sqrt{2}}{\sqrt{\pi}}\int_0^\infty \frac{\sin3\cdot 3}{3}\cos(3 x)e^{-\frac{3^2}{4}t}d3.$$ Does this make any sense? Now it is corrected, but I hope you see via this example why you were wrong. – Gary May 04 '22 at 13:11
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    Did you forget your manners ? This is not the way we work here. – Claude Leibovici May 04 '22 at 13:20
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    In your original version the only mistake was that you had $\lambda$ instead of $x$ in the argument of the error functions. That is what Claude was referring to. – Gary May 04 '22 at 13:20
  • @Gary, fine, I see that. – Luthier415Hz May 04 '22 at 13:22
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    By the way, ou edited after my comment. Look at the time stamps – Claude Leibovici May 04 '22 at 13:30
  • That is because I trusted you, still I didn't understand what you meant. Now you can give it a 10 each? Thanks! – Luthier415Hz May 04 '22 at 13:44

1 Answers1

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In this answer it's shown that $$ \frac{2}{\pi} \int_0^\infty e^{-\left(\frac{x}{2\xi} \right)^2}\frac{\sin(x)}{x} \, \mathrm{d}x = \mathrm{erf}(\xi) \qquad \text{for} \quad \xi \in \mathbb{R}$$ which means $$ \int_{0}^{\infty}\frac{\sin(\alpha u)}{u}e^{-\frac{u^2}{4}t}\mathrm{d}u \overset{\color{blue}{\alpha u \to u}}{=} \int_{0}^{\infty}\frac{\sin(u)}{u} e^{-\left(\frac{u}{2\frac{\alpha}{\sqrt{t}}} \right)^2} \mathrm{d}u =\frac{\pi}{2} \mathrm{erf}\left(\frac{\alpha}{\sqrt{t}}\right),\qquad \text{for} \quad \alpha \in \mathbb{R} \text{ and } t>0 $$ Finally, since $\sin(ax)\cos(bx) = \frac{1}{2}\left[\sin((a-b)x) +\sin((a+b)x)\right]$ you get \begin{align} \int_{0}^{\infty} \frac{\sin(au)\cos(bu)}{u} e^{-\frac{u^2}{4}t}\mathrm{d}u & = \frac{1}{2}\left[\int_{0}^{\infty}\frac{\sin((a-b) u)}{u}e^{-\frac{u^2}{4}t}\mathrm{d}u + \int_{0}^{\infty}\frac{\sin((a+b) u)}{u}e^{-\frac{u^2}{4}t}\mathrm{d}u \right]\\ & = \frac{\pi}{4}\left[\mathrm{erf}\left(\frac{a-b}{\sqrt{t}}\right) + \mathrm{erf}\left(\frac{a+b}{\sqrt{t}}\right) \right] \end{align} as desired.

Robert Lee
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