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Improper integral with a parameter $int_{0}^{infty} e^{-cx^{2}}sin(tx)dx$

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up vote 2 down vote favorite I need to evaluate the following integral. $$int_{0}^{infty} e^{-cx^{2}}sin(tx) ~dx$$ Here's what I've done so far. $$I(t) = int_{0}^{infty} e^{-cx^{2}}sin(tx)~dx$$ Then differentiating under the integral sign I get: $$I'(t) = int_{0}^{infty} xe^{-cx^{2}}cos(tx)~dx$$ This is pretty straightforward and integrating by parts I get this: $$I '(t) =frac{1}{2c}+frac{t}{2c}int_{0}^{infty} e^{-cx^{2}}sin(tx)~dx$$ which can be written as $$I'(t) = frac{1}{2c}+frac{t}{2c}I(t)$$ Rearranging I end up with the following differential equation: $$I'(t)+frac{t}{2c}I(t)=frac{1}{2c}$$ When I try to solve this I end up with a different result than what an online calculator got: $$-dfrac{sqrt{{pi}}mathrm{i}mathrm{e}^{-frac{t^2}{4c}}operatorname{erf}left(frac{mathrm{i}t}{2sqrt{c...

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