**Evaluate**: $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$

**Solution**: $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\infty}e^{-x^2} dx +2 \int_{0}^{\infty}(2x^4)e^{-x^2} dx = \int_{-\infty}^{\infty}e^{-x^2} dx +4\int_{0}^{\infty}x^4e^{-x^2} dx .$$ It is well-known that $$\int_{-\infty}^{\infty}e^{-x^2} dx = \sqrt{\pi}.$$

We will calculate the second integral $$4\int_{0}^{\infty}x^4e^{-x^2}dx.$$

Substitute $u = x^2$ then the integral transforms to $2 \int_{0}^{\infty}u^{\frac{3}{2}}e^{-u}du = 2 \Gamma(\frac{3}{2}) = 2 \cdot \frac{3}{4}\cdot \sqrt{\pi} = \frac{3}{2} \sqrt{\pi}$

Hence, $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \sqrt{\pi} + \frac{3}{2} \sqrt{\pi} = \frac{5}{2} \sqrt{\pi}.$$

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Sir how the integral write into two part

ReplyDelete$x^4$ is an even function. So -a to a integral even function = 2 times 0 to a integral of the function. is it clear? I have edited the answer. plz check. also plz follow me be mail and subscribe.

ReplyDeleteI have updated the solution. Hopefully now its clear. Any doubts, feel free to ask. All the best.

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