# Solving Integral of exp{-(a*x^2+b*x+c)} from -inf to inf

• yf920
In summary, the integral of exp{-(a*x^2+b*x+c)} from 0 to infinite can be calculated by rewriting the exponent as -g(kx+d)^2 + f and then substituting u = g(kx+d). This results in a known integral of the form e^(-u^2). The final result is sqrt(pi/a)*exp((b^2-4ac)/4a).
yf920
Integral of exp{-(a*x^2+b*x+c)} from 0 to infinite

i know the answer of Integral of exp{-(a*x^2+b*x+c)} from -inf to inf
= sqrt(pi/a)*exp((b^2-4ac)/4a)

thanks!

Show some work! Forum policy :-]

Write the exponent in the form -g(kx+d)^2 + f. Put g(kx+d)=u. Then the integral
(-inf to inf) becomes twice the same integral over (0 to inf), since the integrand is
now of the form e^(-u^2) which is symmetric about 0.

@ssd, if you mean that the end result of the integral in the original post is half the result of the other one given in there, you are making a calculation error.

OK yf920, let's calculate the result given. The argument of the exponential function can be written as:

$$-ax^2-bx-c=-a\left(x+\frac{b}{2a}\right)^2 +\frac{b^2-4ac}{4a}$$

Putting this into the integral gives then:

$$I=e^{\frac{b^2-4ac}{4a}}\int_{-\infty}^{\infty} e^{-a\left(x+\frac{b}{2a}\right)^2}dx$$

Putting as ssd said:

$$\sqrt{a}\left(x+\frac{b}{2a}\right)=t$$

gives now:

$$I=\frac{e^{\frac{b^2-4ac}{4a}}}{\sqrt{a}}\int_{-\infty}^{\infty} e^{-t^2}dt$$

The following integral is known:

$$\int_{-\infty}^{\infty} e^{-t^2}dt=\sqrt{\pi}$$

Giving the final result as:

$$I=\sqrt{\frac{\pi}{a}} e^{\frac{b^2-4ac}{4a}}$$

Now, the original question was:

$$I=\int_{0}^{\infty} e^{-\left(ax^2+bx+c\right)}dx$$

Which is not the double of the one just derived. Just go through all the steps with this new integral limits and use the result:

$$erf(x)=\frac{2}{\sqrt{\pi}} \int_{0}^{x}e^{-z^2}dz$$

Do these calculations, and come back with the result you found, it will be our pleasure to check it.

## 1. What is the general formula for solving the integral of exp{-(a*x^2+b*x+c)} from -inf to inf?

The general formula for solving this integral is:∫ exp{-(a*x^2+b*x+c)} dx = √(π/a) * exp{(b^2-4ac)/4a}

## 2. How do you approach the integration of exp{-(a*x^2+b*x+c)} from -inf to inf?

This integral can be solved using the substitution method. Let u = √a * x + b/2√a, then the integral becomes:∫ exp{-u^2} du = √π

## 3. Can the integral of exp{-(a*x^2+b*x+c)} from -inf to inf be solved using integration by parts?

No, integration by parts cannot be used to solve this integral. It requires the substitution method as mentioned in the previous answer.

## 4. Is there a specific value of a, b, and c that makes the integral of exp{-(a*x^2+b*x+c)} from -inf to inf solvable?

Yes, for the integral to be solvable, a must be a positive number and b^2-4ac must be negative. This ensures that the exponential term approaches 0 as x approaches ±∞, making the integral converge.

## 5. What is the significance of the integral of exp{-(a*x^2+b*x+c)} from -inf to inf in mathematics and science?

The integral of exp{-(a*x^2+b*x+c)} from -inf to inf is commonly used in statistics and probability to calculate the normal distribution. It is also used in physics and engineering to solve problems involving Gaussian functions. Additionally, it has applications in quantum mechanics and signal processing.

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