Compute $\int_{1}^{\infty} e^{2x - x^2}\mathrm dx$

Question

I couldn't find the appropriate transformation for this integral. Can someone help me?

$$\int_{1}^{\infty} e^{2x - x^2}\mathrm dx$$

Answer 1

Hints:

• $\mathrm{e} \cdot \frac{1}{\mathrm{e}} = 1$.
• $-1+2x-x^2 = -(x-1)^2$
• Then this is a shifted Gaussian integral, as observed in comments by Sangchul Lee.

Answer 2

Start by completing the square in the exponent:

\begin{align} -x^2+2x&=-(x^2-2x) \\ &=-(x^2-2x+1-1) \\ &=-(x-1)^2+1 \end{align} Your integral is then equal to \begin{align} \int_1^{+\infty} e^{-(x-1)^2+1}\,dx&=e\int_1^{+\infty} e^{-(x-1)^2}\,dx \\ &=e\int_0^{+\infty} e^{-x^2}\,dx \\ &=\boxed{\frac{e\sqrt{\pi}}{2}} \end{align}