Integration of $\int_{0}^{\infty}e^{-(x+x^{2})}dx$

Question

I have the following integral

$$\int\limits_0^{\infty}e^{-(x+x^{2})}dx$$

I am wondering if there is an analytical solution for this integral. Any hints or suggestions?

Answer 1

Complete the square in the exponent and change variables to find:

$$\frac{1}{2} \sqrt[4]{e} \sqrt{\pi } \text{erfc}\left(\frac{1}{2}\right)$$

Answer 2

Well, we have:

$$\mathcal{I}_{\space\text{n}}\left(\text{s}\right):=\int_0^\infty\exp\left(-\left(\text{s}\cdot x+x^\text{n}\right)\right)\space\text{d}x\tag1$$

Now, for the $\exp$-function we can write:

$$\exp\left(-\left(\text{s}\cdot x+x^\text{n}\right)\right)=\exp\left(-\text{s}\cdot x-x^\text{n}\right)=\exp\left(-\text{s}\cdot x\right)\cdot\exp\left(-x^\text{n}\right)\tag2$$

So, we get:

$$\mathcal{I}_{\space\text{n}}\left(\text{s}\right)=\int_0^\infty\exp\left(-\text{s}\cdot x\right)\cdot\exp\left(-x^\text{n}\right)\space\text{d}x\tag3$$

Which is the Laplace transform of $\exp\left(-x^\text{n}\right)$:

$$\mathcal{I}_{\space\text{n}}\left(\text{s}\right)=\mathcal{L}_x\left[\exp\left(-x^\text{n}\right)\right]_{\left(\text{s}\right)}\tag4$$

Using:

$$\exp\left(-x^\text{n}\right)=\sum_{\text{k}=0}^\infty\frac{\left(-x^\text{n}\right)^\text{k}}{\text{k}!}=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot x^{\text{n}\cdot\text{k}}\tag5$$

We can rewrite equation $\left(4\right)$:

$$\mathcal{I}_{\space\text{n}}\left(\text{s}\right)=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\mathcal{L}_x\left[x^{\text{n}\cdot\text{k}}\right]_{\left(\text{s}\right)}=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{\Gamma\left(1+\text{n}\cdot\text{k}\right)}{\text{s}^{1+\text{n}\cdot\text{k}}}\tag6$$

Answer 3

$$I=\int_{0}^{\infty}{e^{-(x+x^2)}dx}=\int_{0}^{\infty}{e^{-[(x+0.5)^2-0.25]}dx}=e^\frac{1}{4}\int_{0}^{\infty}{e^{-(x+0.5)^2}dx}$$ $u=x+\frac{1}{2}, du=dx$ $$\therefore I=e^\frac{1}{4}\int_{\frac{1}{2}}^{\infty}{e^{-u^2}du}=e^\frac{1}{4}\left(\frac{\sqrt{\pi}}{2}-\int_{0}^{\frac{1}{2}}{e^{-u^2}du}\right)$$ And I think the final integral at the end can maybe expressed in terms of $erf(x)$ or $erfc(x)$