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I am trying to integrate the following:

$$\int e^{-2y}\cos(y^2) \, dy$$

I cannot identify a suitable substitution, and integration by parts would seem to go round in circles.

Please provide me with any minor hints in the right direction (homework).

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Are you sure you have the problem right? You may know that if $f(x)$ is a quadratic polynomial then there is no closed form for $\int e^{f(x)}\,dx$, and that's what your integral looks like. –  Gerry Myerson Oct 15 '12 at 0:39

2 Answers 2

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Is it given as indefinite integral as you wrote? If it is a definite (improper) integral given from $-\infty$ to $\infty$, I would try by writing $e^{iy^{2}} = \cos(y^{2}) + i\sin(y^{2})$. Convergence seems easy to show in this case.

I don't know if it is doable for indefinite integral.

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Notice $\int{e^{-2y}cos(y^2)}dy = -1/2\int{cos(y^2)de^{-2y}}$. Now use Integration by parts. It is not suppose to go round on circles. Just Let $I = \int{e^{-2y}cos(y^2)}dy$, and then until you get the same expression back, you just solve for $I$.

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The problem is that after using IBP twice, you get $\int y^2e^{-2y} \cos(y^2) $... –  N. S. Oct 15 '12 at 0:31

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