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Let $$F(y)=\int_0^\infty e^{-2x}\cos(2xy)\, dx$$ for $y\in \mathbb{R}$. Show that $F$ satisfies $F'(y)+2yF(y)=0$.

I've shown that I can differentiate under the integral sign, and this gives $F'(y)=-2\int_0^\infty xe^{-2x}\sin(2xy)\,dx$, but how in the world is this equal to $-2yF(y)$? Is there some trick I'm missing here?

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I haven't tried, but integration by parts? –  Tunococ Aug 22 '12 at 23:52
    
Probably you should try integrating by parts. Since in that case if you differentiate with respect to $x$, a $y$ factor would come out of the $\sin{(2xy)}$, and $xe^{-2x}$ can be integrated easily. –  Adrián Barquero Aug 22 '12 at 23:55
    
There is a mistake in the question you posted. See the answer. –  Mhenni Benghorbal Aug 23 '12 at 7:30

2 Answers 2

up vote 4 down vote accepted

Integration by parts: $$u=\sin 2xy\,\,,\,\,u'=2y\cos 2xy$$ $$v'=-xe^{-2x}\,\,,\,\,v=\frac{1}{2}e^{-2x}\,\,\,,\,\,\text{so:}$$

$$\left.-2\int_0^\infty xe^{-2x}\sin 2xy\,\,dx=\sin 2xy\,\,e^{-2x}\right|_0^\infty-2y\int_0^\infty e^{-2x}\cos 2xy\,\,dx$$

Now just check the first term in the RHS above is zero and you're done.

Added: The above is wrong and the reason why is in the comment by Adrian below, yet the solution is way more involved than it'd appear and I'll try to post it later.

Further added: I already tried but I can't get what the OP asks. Putting: $$\mathcal I:=\int_0^\infty xe^{-2x}\sin 2xy\,\,dx\,\,\,,\,\,\,\mathcal J:=\int_0^\infty e^{-2x}\cos 2xy\,\,dx$$ I get the following integrating by parts (first time $\,u=x\sin 2xy\,$ , second time $\,\sin 2xy\,$ , third time $\,u=x\cos xy\,$ , and all the times $\,v'=e^{-2x}\,$ ):

$$\mathcal I=\stackrel{\text{this equals zero}}{\overbrace{\left.-\frac{1}{2}xe^{-2x}\sin 2xy\right|_0^\infty}}+\frac{1}{2}\int_0^\infty e^{-2x}\sin 2xy\,\,dx+y\int_0^\infty xe^{-2x}\cos 2xy\,\,dx\Longrightarrow $$ $$\Longrightarrow \left.\mathcal I=-\frac{1}{4}e^{-2x}\sin 2xy\right|_0^\infty+\frac{y}{2}\mathcal J-\left.\frac{y}{2}xe^{-2x}\cos 2xy\right|_0^\infty+\frac{y}{2}\mathcal J-y^2\mathcal I$$

as the limits above equal zero, we get: $$\mathcal I=yJ-y^2\mathcal I\Longrightarrow \mathcal I=\frac{y}{y^2+1}J$$ so I get that denominator $\,y^2+1\,$ there that shouldn't, according to the OP, be there. Either I'm mistaken or the OP is, but I already checked this several times.

Everybody is wholeheartedly invited to either confirm my calculations of refute them. Thanks.

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Dear DonAntonio, just a small correction. Your antiderivative for $v' = -xe^{-2x}$ is incorrect. It should be $\frac{1}{2}xe^{-2x} + \frac{1}{4}e^{-2x}$. But that would not change the fact that the corrected first term in the RHS of your identity would be zero. –  Adrián Barquero Aug 23 '12 at 0:09
    
You're of course right, @AdriánBarquero. For some reason I saw that $\,x\,$ as a $\,1\,$. Thanks a lot and I shall correct the post. –  DonAntonio Aug 23 '12 at 1:47
    
I think you are right. This is distressing: this question came from an analysis qualifying exam a few years ago.... –  Bey Aug 23 '12 at 21:44
1  
Don't worry about that, @Bey : it wouldn't be the first time, nor will it be the last one, that some serious mistake is discovered in some book, exam or whatever. Anyway, I'd try to check this with the instructor/lecturer who wrote the exam or the answers to it: it could be perfectly well I am the wrong one. –  DonAntonio Aug 24 '12 at 2:17
    
@Don: Thanks, I'll do that! –  Bey Aug 24 '12 at 2:31

There is a mistake in the question posted. First note that, if we consider the integral

$$ \int_0^\infty e^{-sx}\cos(2xy)\, dx = {\frac {s}{{s}^{2}+4\,{y}^{2}}} $$

which gives the Laplace transform of $ \cos(2xy) $. Substituting $s=2$ gives

$$ F(y) = {\frac {2}{{4}+4\,{y}^{2}}} $$

Now, there is no way that $F(y)$ satisfies the given differential equation, trying this yields

$$ F'(y)+2yF(y)={\frac {{y}^{3}}{ \left( 1+{y}^{2} \right) ^{2}}} \neq 0 \,. $$

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