How the limit of a definite integral affected the integral value?

Question

Consider the integral $f(x)=\int_{-\infty }^{\infty}xe^{x^{2}}$
To determine the value of this integral we can substitute $x^{2}=u$ so the integrand is reduced to $\frac{e^{u}}{2}$.
Again resubstitution $x^{2}=u$ gives us $f(x)=\frac{e^{x^{2}}}{2}$. Now if we substitute the limits $x\rightarrow \infty$ and $x\rightarrow -\infty$ we get the value of the integral $f(x)=\int_{-\infty }^{\infty}xe^{x^{2}}\rightarrow \infty$
We can also calculate the reduced integral $\frac{e^{u}}{2}$ with limits $u\rightarrow\infty$ to $u\rightarrow\infty$ that is $\int_{\infty }^{\infty}\frac{-e^{u}}{2}=0$.
The actual answer using the property of an odd function is $f(x)=\int_{-\infty }^{\infty}xe^{x^{2}}=0$
I am not getting what integration property is violated here.

Answer 1

This is a case of a principal value for an integral.

You get $$I_R=\int_{-R}^R xe^{x^2}\,dx =0$$ because $f(x)=xe^{x^2}$ is odd: $f(-x)=-f(x).$ So $$\lim_{R\to+\infty}I_R=0.$$

But $$\lim_{(R,S)\to(+\infty,+\infty)}\int_{-R}^S xe^{x^2}\,dx$$ is undefined.

Your substitution essentially is like taking $\int_{-R}^R.$

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You can do the same for $$\int_{-\infty}^{\infty} x\,dx,\tag1$$ letting $u=x^2$ and get $$\int_{\infty}^{\infty}\frac12\,du=0$$ but (1) still doesn’t converge when $R,S$ go to infinity separately.

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One reason we tend to reject principal values is that $$\int_{-\infty}^\infty f(x)\,dx$$ has a different principal value than $$\int_{\infty}^\infty f(x+a)\,dx\tag2$$ (this also means principal values don’t work with substitution, because $u=x+a$ doesn’t work in (2).)

Answer 2

I just noticed your edit, and this is terrible notation. The result after a definite integral of a single-variable function is NEVER a function again. If you have a definite integral your result is a number.

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First of all, whenever we write down the symbol $\int_{-\infty}^{\infty}f(x)\,dx$, the standard meaning is the following (working with Riemann integrals throughout):

• we have a function $f:\Bbb{R}\to\Bbb{R}$ such that for any $a,b\in\Bbb{R}$ with $a<b$, we have that the restriction $f:[a,b]\to\Bbb{R}$ is (properly) Riemann-integrable. This means $f$ is bounded on $[a,b]$ and the supremum of lower sums equals infimum of upper sums.
• The limit $\lim\limits_{r\to \infty}\int_0^{r}f(x)\,dx$ exists in $\Bbb{R}$
• The limit $\lim\limits_{s\to -\infty}\int_{s}^{0}f(x)\,dx$ exists in $\Bbb{R}$.

In more common parlance, the second and third conditions express that the limit must exist and be finite. In this case, we say $f$ is improperly Riemann-integrable on $\Bbb{R}$ and we define \begin{align} \int_{-\infty}^{\infty}f(x)\,dx := \lim\limits_{r\to \infty}\int_0^{r}f(x)\,dx+ \lim\limits_{s\to -\infty}\int_{s}^{0}f(x)\,dx. \end{align}

If one intends ANY OTHER meaning, that should be clearly stated in words prior to writing down this symbol. In your case, the function $f(x)=xe^{x^2}$ DOES NOT satisfy this condition (because the two limits are $\infty$ due to the $e^{x^2}$ term), so there is no meaning to writing down $\int_{-\infty}^{\infty}xe^{x^2}\,dx$. If you write this down as it is, then it has as much meaning as me writing \begin{align} \ddot{\smile}, *,\text{spongebob}+ \text{power rangers}= \text{caramel ice-cream}. \end{align} In some cases, one can work with "symmetric limits", and consider $\lim\limits_{r\to \infty}\int_{-r}^rf(x)\,dx$. For the function $f(x)=xe^{x^2}$, this limit is $0$ because you're integrating an odd function over a symmetric interval. You can also see this by a direct substitution. We define $f(x)=xe^{x^2}$ and $g(x)=x^2$ and $\phi(x)=e^x$. Then, for any $r>0$, we have \begin{align} \int_{-r}^rf(x)\,dx&=\int_{-r}^r\phi(g(x))\cdot\frac{g'(x)}{2}\,dx\\ &=\frac{1}{2}\int_{g(-r)}^{g(r)}\phi(t)\,dt\\ &=\frac{1}{2}\int_{r^2}^{r^2}e^t\,dt\\ &=0. \end{align} So, as long as you're dealing with the symmetric limit, then the integral is $0$. But you have to mention this explicitly because otherwise, by definition the function $f$ is NOT improperly Riemann-integrable.