1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ The derivatives bring down $4/a^{2}$, not just 4. $\endgroup$ – Ininterrompue Jan 11 at 5:17
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.