integrating $\int\limits_{ - \infty }^\infty {\sin ax{e^{ - {x^2}}}dx} $

Question

I want to solve the following equation:

$${u_t} = {u_{xx}} + {u_{yy}} + {e^t}$$ $$u\left( {x,y,0} \right) = \cos x\sin y$$

General form of equation: $${u_t} = {a^2}\left( {{u_{xx}} + {u_{yy}}} \right) + f\left( {x,y,t} \right)$$ $$u\left( {x,y,0} \right) = \varphi \left( {x,y} \right)$$

The solution of the general form: $$u\left( {x,y,t} \right) = \frac{1}{{4{a^2}\pi t}}\int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {\varphi \left( {\xi ,\eta } \right){e^{ - \frac{{{{\left( {x - \xi } \right)}^2} + {{\left( {y - \eta } \right)}^2}}}{{4{a^2}t}}}}} d\xi d\eta } + \int\limits_0^t {\int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {\frac{{f\left( {\xi ,\eta } \right)}}{{4{a^2}\pi \left( {t - \tau } \right)}}{e^{ - \frac{{{{\left( {x - \xi } \right)}^2} + {{\left( {y - \eta } \right)}^2}}}{{4{a^2}\left( {t - \tau } \right)}}}}} d\xi d\eta } } d\tau $$

Applying the formula: $$u\left( {x,y,t} \right) = \frac{1}{{4\pi t}}\int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {\cos \xi \sin \eta {e^{ - \frac{{{{\left( {x - \xi } \right)}^2} + {{\left( {y - \eta } \right)}^2}}}{{4t}}}}d\xi d\eta } } + \frac{1}{{4\pi }}\int\limits_0^t {\frac{{{e^\tau }d\tau }}{{t - \tau }}\int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {{e^{ - \frac{{{{\left( {x - \xi } \right)}^2} + {{\left( {y - \eta } \right)}^2}}}{{4t}}}}d\xi } d\eta } } $$

After simplifying: $$u\left( {x,y,t} \right) = \frac{1}{\pi }\int\limits_{ - \infty }^\infty {\sin \left( {y + 2\sqrt t {s_1}} \right){e^{ - s_1^2}}d{s_1}} \int\limits_{ - \infty }^\infty {\cos \left( {x + 2\sqrt t {s_2}} \right){e^{ - s_2^2}}d{s_2}} + {e^t} - 1$$ My question is how do I integrate the last integrals?

Answer 1

For $$ S := \int_{-\infty}^{\infty} \sin(b+as)e^{-s^2}\,ds\qquad\text{and}\qquad C := \int_{-\infty}^{\infty} \cos(b+as)e^{-s^2}\,ds , $$ we can do this: $S = \operatorname{Im} B$ and $C = \operatorname{Re} B$ where $$ B := \int_{-\infty}^{\infty} \exp\big(-s^2+i(b+as)\big)\;ds . $$ Complete the square \begin{align} B &= \int_{-\infty}^{\infty} \exp\big(-(s-\frac{ia}{2})^2+(-\frac{a^2}{4}+ib)\big)\;ds \\&= \exp\big(-\frac{a^2}{4}+ib\big)\int_{-\infty}^{\infty} \exp\big(-(s-\frac{ia}{2})^2\big)\;ds \\&= \exp\big(-\frac{a^2}{4}+ib\big)\sqrt{\pi} . \end{align} Thus $$ S = \sqrt{\pi}\sin(b)\exp\big(-\frac{a^2}{4}\big) \qquad\text{and}\qquad C = \sqrt{\pi}\cos(b)\exp\big(-\frac{a^2}{4}\big) $$

Answer 2

Compute instead $$I=\int e^{-s^2+i \left(2 s \sqrt{t}+y\right)}\,ds$$ Complete the square to face a well know integrals, making $$I=\frac{\sqrt{\pi }}{2}\, e^{-t+i y}\, \text{erf}\left(s-i \sqrt{t}\right)$$

The question now is : do the real and imaginary parts converge for infinite bounds ?

Answer 3

We try to evaluate $$f(\alpha,\beta)=\int_{-\infty}^\infty e^{-x^2}\cos(\alpha x+\beta)dx.$$ Note that $$ f(\alpha,\beta){=\int_{-\infty}^\infty e^{-x^2}\cos(\alpha x+\beta)dx \\= \int_{-\infty}^\infty e^{-x^2}[\cos\alpha x\cos\beta-\sin\alpha x\sin\beta] dx \\= \cos\beta\int_{-\infty}^\infty e^{-x^2}\cos\alpha x dx . } $$ Using integration by parts for $g(\alpha)=\int_{-\infty}^\infty e^{-x^2}\cos\alpha x dx$, we obtain $$ \int_{-\infty}^\infty e^{-x^2}\cos\alpha x dx{= \frac{1}{\alpha}e^{-x^2}\sin\alpha x\Big|_{-\infty}^\infty+\frac{2}{\alpha}\int_{-\infty}^\infty xe^{-x^2}\sin\alpha x dx \\= \frac{2}{\alpha}\int_{-\infty}^\infty xe^{-x^2}\sin\alpha x dx \\= -\frac{2}{\alpha}g'(\alpha). } $$ Hence $$ \frac{g'(\alpha)}{g(\alpha)}=-\frac{\alpha}{2}\implies g(\alpha)=Ce^{-\frac{\alpha^2}{4}}, $$ where $C=g(0)=\int_{-\infty}^\infty e^{-x^2}dx=\sqrt\pi$. Finally, $$ \int_{-\infty}^\infty e^{-x^2}\cos(\alpha x+\beta)dx=\sqrt\pi e^{-\frac{\alpha^2}{4}}\cos\beta. $$