My Math Forum integrals

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 January 8th, 2012, 11:29 AM #1 Member   Joined: Jan 2012 Posts: 86 Thanks: 0 integrals $\int_{0}^{\infty} \sin (ax^2) \ \cos (2bx) \ dx$ $\int_{0}^{\infty} \cos (ax^2) \ \cos(2bx) \ dx$ I evaluated $\int_{0}^{\infty} e^{-ax^{2}} \ \cos (2bx) \ dx$ (which can be done in numerous ways) and then tried to argue that the answer is valid for imaginary $a$ by using the uniqueness theorem and Morera's theorem from complex analysis. Unfortunately I got nowhere with that. What's another way to evaluate these integrals?
 January 8th, 2012, 02:56 PM #2 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: integrals Ok I'd go with: $\operatorname{Im} \left\{ {\int\limits_0^\infty {{e^{ia{x^2}}}\cos \left( {2bx} \right)dx} } \right\}= \int\limits_0^\infty {\sin \left( {a{x^2}} \right)\cos \left( {2bx} \right)dx}$ (same for $Re$) And consider like you an arbitrary complex number $z$ $\int\limits_0^\infty {{e^{z{x^2}}}\cos \left( {2bx} \right)dx}=$ ${x^2}= - u$ $dx= \frac{{idu}}{{2\sqrt u }}$ $\frac{i}{2}\int\limits_0^\infty {{e^{ - zu}}\frac{{\cosh \left( {2b\sqrt u } \right)}}{{\sqrt u }}du}= \frac{i}{2}\mathcal{L}\left( {\frac{{\cosh \left( {2b\sqrt u } \right)}}{{\sqrt u }}} \right)\left( z \right)$ Solving this with Wolfram I get that the expression is: $\frac{i}{2}\sqrt {\frac{\pi }{z}} {e^{\frac{{{b^2}}}{z}}}$ Now plug in z = $ai$ to get $\frac{{\sqrt i }}{2}\sqrt {\frac{\pi }{a}} {e^{ - i\frac{{{b^2}}}{a}}}$ $\frac{{i + 1}}{{2\sqrt 2 }}\sqrt {\frac{\pi }{a}} \left( {\cos \left( {\frac{{{b^2}}}{a}} \right) - i\sin \left( {\frac{{{b^2}}}{a}} \right)} \right)$ And taking $Im$ and $Re$ we get that the integrals are $\int\limits_0^\infty {\sin \left( {a{x^2}} \right)\cos \left( {2bx} \right)dx}= \sqrt {\frac{\pi }{{8a}}} \left\{ {\cos \left( {\frac{{{b^2}}}{a}} \right) - \sin \left( {\frac{{{b^2}}}{a}} \right)} \right\}$ $\int\limits_0^\infty {\cos \left( {a{x^2}} \right)\cos \left( {2bx} \right)dx}= \sqrt {\frac{\pi }{{8a}}} \left\{ {\cos \left( {\frac{{{b^2}}}{a}} \right) + \sin \left( {\frac{{{b^2}}}{a}} \right)} \right\}$ I do not know how far this is correct but do it yourself and check if there is something wrong. The case $a=1$ and $b=0$ gives the correct Fresnel integral values: $\int\limits_0^\infty {\sin \left( {{x^2}} \right)dx}= \int\limits_0^\infty {\cos \left( {{x^2}} \right)dx} = \sqrt {\frac{\pi }{8}}$
 January 8th, 2012, 03:37 PM #3 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: integrals I'll illustrate how to solve the Laplace Transform (so I don't leave holes in the solution): $\int\limits_0^\infty {{e^{ - zu}}\frac{{\cosh \left( {2b\sqrt u } \right)}}{{\sqrt u }}du}=$ $\int\limits_0^\infty {{e^{ - zu}}\frac{{{e^{2b\sqrt u }} + {e^{ - 2b\sqrt u }}}}{{2\sqrt u }}du}=$ Change the variable $\sqrt u= m$ $\int\limits_0^\infty {{e^{ - zu}}\frac{{{e^{2b\sqrt u }}}}{{2\sqrt u }}du} + \int\limits_0^\infty {{e^{ - zu}}\frac{{{e^{ - 2b\sqrt u }}}}{{2\sqrt u }}du}= \int\limits_0^\infty {{e^{2mb - z{m^2}}}dm} + \int\limits_0^\infty {{e^{ - z{m^2} - 2mb}}dm}$ Express the polinomials as: $2mb - z{m^2}= - z\left( {{m^2} - 2\frac{{mb}}{z} + \frac{{{b^2}}}{{{z^2}}} - \frac{{{b^2}}}{{{z^2}}}} \right) = - z{\left( {m - \frac{b}{z}} \right)^2} + \frac{{{b^2}}}{z}$ $- 2mb - z{m^2}= - z\left( {{m^2} + 2\frac{{mb}}{z} + \frac{{{b^2}}}{{{z^2}}} - \frac{{{b^2}}}{{{z^2}}}} \right) = - z{\left( {m + \frac{b}{z}} \right)^2} + \frac{{{b^2}}}{z}$ ${e^{\frac{{{b^2}}}{z}}}\left\{ {\int\limits_0^\infty {{e^{ - z{{\left( {m - \frac{b}{z}} \right)}^2}}}dm} + \int\limits_0^\infty {{e^{ - z{{\left( {m + \frac{b}{z}} \right)}^2}}}dm} } \right\}$ Change the variable in the first integral $m= -u$ ${e^{\frac{{{b^2}}}{z}}}\left\{ { - \int\limits_0^{ - \infty } {{e^{ - z{{\left( { - u - \frac{b}{z}} \right)}^2}}}du} + \int\limits_0^\infty {{e^{ - z{{\left( {m + \frac{b}{z}} \right)}^2}}}dm} } \right\}$ ${e^{\frac{{{b^2}}}{z}}}\left\{ {\int\limits_{ - \infty }^0 {{e^{ - z{{\left( {u + \frac{b}{z}} \right)}^2}}}du} + \int\limits_0^\infty {{e^{ - z{{\left( {m + \frac{b}{z}} \right)}^2}}}dm} } \right\}= {e^{\frac{{{b^2}}}{z}}}\int\limits_{ - \infty }^\infty {{e^{ - z{{\left( {m + \frac{b}{z}} \right)}^2}}}dm}$ Change the variable $m+\frac{b}{z}= x$ ${e^{\frac{{{b^2}}}{z}}}\int\limits_{ - \infty }^\infty {{e^{ - z{m^2}}}dm}=$ One last variable change $\sqrt z m= x$ ${e^{\frac{{{b^2}}}{z}}}\int\limits_{ - \infty }^\infty {{e^{ - z{m^2}}}dm}= {e^{\frac{{{b^2}}}{z}}}\frac{1}{{\sqrt z }}\int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}x} = \sqrt {\frac{\pi }{z}} {e^{\frac{{{b^2}}}{z}}}$
 January 8th, 2012, 04:14 PM #4 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: integrals $\int_{0}^{\infty} \sin(ax^2) \ \cos(2bx) \ dx$ Let us first rewrite $\sin(ax^2) \ \cos(2bx)$ $\sin (ax^2)cos(2bx)=\frac{1}{2} \left{ \sin (ax^2+2bx) + \sin (ax^2-2bx) \right}$ $=\frac{1}{2} \left{ \sin (a(x+\frac{b}{a})^2-\frac{b^2}{a}) + \sin (a(x-\frac{b}{a})^2-\frac{b^2}{a}) \right}$ $=\frac{1}{2} \left{ \sin (a(x+\frac{b}{a})^2) \cos(\frac{b^2}{a}) - \cos(a(x+\frac{b}{a})^2) \sin(\frac{b^2}{a}) + \sin (a(x-\frac{b}{a})^2) \cos(\frac{b^2}{a}) - \cos(a(x-\frac{b}{a})^2) \sin(\frac{b^2}{a}) \right}$ $\int_{0}^{\infty} \sin(ax^2) \ \cos(2bx) \ dx$ $=\frac{1}{2} \left{ \cos(\frac{b^2}{a}) \int_{0}^{\infty} \sin (a(x+\frac{b}{a})^2) + \sin (a(x-\frac{b}{a})^2) \ dx - \sin(\frac{b^2}{a}) \int_{0}^{\infty} \sin (a(x+\frac{b}{a})^2) + \sin (a(x-\frac{b}{a})^2) \ dx \right}$ $=\frac{1}{2a} \left{ \cos(\frac{b^2}{a}) \int_{0}^{\infty} \sin (a(x+\frac{b}{a})^2) + \sin (a(x-\frac{b}{a})^2) \ dax - \sin(\frac{b^2}{a}) \int_{0}^{\infty} \sin (a(x+\frac{b}{a})^2) + \sin (a(x-\frac{b}{a})^2) \ dax \right}$ $=\frac{1}{2a} \left{ \cos(\frac{b^2}{a}) 2S(\infty) - \sin(\frac{b^2}{a}) 2C(\infty) \right}$ with S and C the Fresnel integrals. $=\frac{1}{2a} \left{ \cos(\frac{b^2}{a}) 2\frac{\sqrt{\pi}}{8} - \sin(\frac{b^2}{a}) 2\frac{\sqrt{\pi}}{8} \right}$ $\int\limits_0^\infty {\cos \left( {a{x^2}} \right)\cos \left( {2bx} \right)dx}= \sqrt {\frac{\pi }{{8a}}} \left\{ {\cos \left( {\frac{{{b^2}}}{a}} \right) + \sin \left( {\frac{{{b^2}}}{a}} \right)} \right\}$ Yes, same result as Weiler!!!
 January 8th, 2012, 05:21 PM #5 Member   Joined: Jan 2012 Posts: 86 Thanks: 0 Re: integrals Thanks. I've posted these on other forums, but no one could offer much help. Weiler You evaluated that Laplace transform as z were a real parameter. Is that justified?
 January 8th, 2012, 05:37 PM #6 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: integrals What steps are you talking about? I solved the integral pretty much with integral theorems and considering $z$ to be complex. Btw, nice one wnvl, trigonometry always helps. You used $\sin a + \sin b= 2\sin \frac{a+b}{2} \cos \frac{a+b}{2}$ , right? What I don't understand is the notation $dax$. It means you changed variables?
 January 8th, 2012, 05:53 PM #7 Senior Member   Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0 Re: integrals wnvl, just realized you solution and mine are the same only you worked in the real line and I worked in the complex plane. (See how if you take Re and Im in my steps it is pretty much the same.)
 January 8th, 2012, 07:02 PM #8 Member   Joined: Jan 2012 Posts: 86 Thanks: 0 Re: integrals I'm not sure what the notation $dax$ means and I'm assume he didn't mean for those two cosine functions to become sine functions when he rewrote the integral, but if you integrate over $(-\infty, \infty)$ instead of $(0,\infty)$ simple substitutions at the end lead to the answer.
January 9th, 2012, 03:08 AM   #9
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Re: integrals

Quote:
 Originally Posted by Weiler What steps are you talking about? I solved the integral pretty much with integral theorems and considering $z$ to be complex. Btw, nice one wnvl, trignometry always helps. You used $\sin a + \sin b= 2\sin \frac{a+b}{2} \cos \frac{a+b}{2}$ , right?
Yes.

Quote:
 Originally Posted by Weiler What I don't understand is the notation $dax$. It means you changed variables?
Yes, I am doing in fact a substitution u=ax.

 January 9th, 2012, 03:31 AM #10 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: integrals I first tried to solve it using complex contour integration, but that didn't work without doing first the goniometric simplifications.

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