In my book, it is said that $$\int e^{-x^2} \, \mathrm{d}x$$ cannot be solved by the method of inspection. It then turned to method of substitution as a new topic. I am not able to solve this expression by method of substitution either. Can someone help me? Thanks.
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may be that expression may not be clearly understandable please ask where it is not. – Sikander Jun 29 '15 at 17:53
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1It can be shown that the function in question has no elementary antiderivative (i.e., it cannot be expressed as a polynomial, logarithm, exponential, trigonometric, etc.) – Clayton Jun 29 '15 at 17:56
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1The definite integral, $\int_{-\infty}^{\infty}e^{-x^2}dx$, can be solved by integrating over the plane in polar coordinates. It is a classic problem. – Darrell Jun 29 '15 at 18:01
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possible duplicate of Is there really no way to integrate $e^{-x^2}$? – Jonas Meyer Jun 30 '15 at 02:36
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Gentlemen I only want to ask whether this expression can be integrated by method of substitution.When my book stated that the method of inspection is not completely reliable it gave this expression as an example.Then it went to other topics like substitution method,integration by parts etc. but it really didn't mentioned whether this expression now can be integrated by these methods.I don't know why I feel that this may be integrated,for my book then must have given some other example which cannot be integrated by inspection method. – Sikander Jun 30 '15 at 16:26
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I am to good extent satisfied with given answers. – Sikander Jun 30 '15 at 18:51
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This antiderivative is not solvable by any of the methods of basic calculus, including the method of substitution. This is proven using the methods of a field of math called differential Galois Theory. However, that doesn't stop mathematicians from inventing a function to solve this. This function is the error function, denoted $\text{erf}$, and is applied in this way:
$$\int e^{-x^2}dx=\frac{\sqrt{\pi}}{2}\text{erf}(x).$$
When I first learned this, I was highly dissatisfied with this answer, but this is (depending on who you ask) the definition of the error function, and its sole purpose is to solve this type of integral.
Plutoro
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This is not an integral with a nice form.
Note that this doesn't mean that all definite versions of this integral don't have nice solutions: for example, $$\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi};$$ this is known as the Gaussian integral.
See https://en.wikipedia.org/wiki/Gaussian_integral for further info.
Noah Schweber
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The integral does not have an elementary anti derivative. That is, it can't be expressed in terms of "simple" functions such as logarithms, exponentials, trigonometrical functions, etc... We have, however that:
$$\int e^{-x^2}\, \mathrm{d}x=\frac{\sqrt{\pi}}{2}\text{erf}(x)$$
where $\text{erf}(x)$ is the error function. We also have a quite a few nice definite integrals, such as the Gaussian Integral:
$$\int_{-\infty}^{\infty} e^{-x^2}\, \mathrm{d}x=\sqrt{\pi}$$
And since the integral is symmetric, we have also
$$\int_{0}^{\infty} e^{-x^2}\, \mathrm{d}x=\frac{\sqrt{\pi}}{2}$$
Zain Patel
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The integral in question does not have a closed form expression in terms of elementary functions. You could expand the integral into $$\int e^{-x^2} {dx}=\int \left(\sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{k!}\right) dx$$ And from there obtain a power-series representation of the indefinite integral. This also has the benefit of being convergent for all real numbers $x$. The first few terms in the series representation of the integral are$$C+x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}...$$Where C is an arbitrary constant of integration.
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$e^{-x^2}$ does not have an elementary antiderivative. In other words, there is no 'simple' function that this integral is equal to.
The integral
$$\int_{-\infty}^{\infty}e^{-x^2}\ dx$$
on the contrary, can be evaluated by polar coordinates with double integrals.
ireallydonknow
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This integral is definitively the Error function and no it cannot be evaluated by using substitution. However, by setting certain limits, for instance \begin{equation}I=\int_0^{\infty}e^{-x^2}dx \end{equation} the integral can be solved (by means of the technique called differentiation under the integral sign) to achieve special values.
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This is indeed a very poor choice for an illustration, as no substitution can help in this case (as other respondents say, this antiderivative is hopeless.)
A better example could be
$$\int xe^{-x^2}\,dx.$$
If you substitute $t=-x^2$, you have that $dt=-2x\,dx$ and you can write
$$\int xe^{-x^2}\,dx=-\frac12\int e^t\,dt.$$
This is an easy antiderivative that gives
$$-\frac{e^t}2+C=-\frac{e^{-x^2}}2+C.$$
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I am learning just introductions but will soon understand till gaussian integral – Sikander Jun 29 '15 at 18:17
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This is an antiderivative problem, also called indefinite integral. The definite integrals that cannot be solved from indefinite ones as an application of the Fundamental Theorem of Calculus can sometimes be solved by other, pretty different techniques. This is a more advanced topic. – Jun 29 '15 at 18:25