# Is this way for integrating has any fault?

I have tried this $$\int_{-\infty}^{\infty}e^{-2|x|/a}\cdot \frac{d^2}{dx^2}(e^{-2|x|/a})~~dx$$ integration in the method below- $$\int_{-\infty}^{\infty}e^{-2|x|/a}\cdot \frac{d^2}{dx^2}(e^{-2|x|/a})~~dx\\=\int_{-\infty}^{0}e^{-2(-x)/a}\cdot \frac{d^2}{dx^2}(e^{-2(-x)/a})~~dx +\int_{0}^{\infty}e^{-2x/a}\cdot \frac{d^2}{dx^2}(e^{-2x/a})~~dx\\ =\int_{-\infty}^{0}e^{2x/a}\cdot \frac{d^2}{dx^2}(e^{2x/a})~~dx+\int_{0}^{\infty}e^{-2x/a}\cdot \frac{d^2}{dx^2}(e^{-2x/a})~~dx\\ =4\int_{-\infty}^{0}e^{4x/a}~~dx+4\int_{0}^{\infty}e^{-4x/a}~~dx\\ =4\cdot \frac{a}{4}+4\cdot \frac a4=2a$$ I am not sure I am correct or not. Please help me with this integral and correct me if I am wrong at any step. Any help will be appreciated.

• The derivatives bring down $4/a^{2}$, not just 4. – Ininterrompue Jan 11 at 5:17

Let $$a>0$$. Function $$f=e^{-2|x|/a}\cdot \frac{d^2}{dx^2}(e^{-2|x|/a})$$ is even. Then $$\int_{-\infty}^{\infty}f\,dx=2\int_{0}^{\infty}f\,dx\\ =2\int_{0}^{\infty}e^{-2x/a}\cdot \frac{d^2}{dx^2}(e^{-2x/a})\,dx\\ =2\int_{0}^{\infty}\frac{4 {e^{-\frac{4 x}{a}}}}{{{a}^{2}}}dx=\frac2a$$