Taking Fourier transform with WolframAlpha gives unexpected result

Question

I have noticed that WolframAlpha's definition of the Fourier Transform is different than the one in my book of formulas. In my book the Fourier transform of $f(t)$ is defined as: $$\int_{-\infty}^{\infty} f(t)e^{- i \omega t}dt$$ The inverse transform is defined as: $$\frac{1}{2 \pi}\int_{-\infty}^{\infty} F(\omega)e^{i \omega t}d\omega$$ Using WolframAlpha with a simple case is shown here. This looks more like the inverse Fourier transform, except we divide by $\sqrt{2 \pi}$ instead of $2 \pi$.

Trying this simple case I could get the same answer as in my table of fourier transforms by multiplying WolframAlpha's answer with $\sqrt{2 \pi}$.

Will I generally get the right result by taking WolframAlphas answer and multiplying by $\sqrt{2 \pi}$? (Since WolframAlpha also integrates in the other direction). Is there a way to get the right (according to definiton above) answer right away using WolframAlpha?