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I'm having lots of troubles while trying to find the second order derivatives of this integral... I don't think it admits a solution...

$$\int_{-\infty}^{\frac{x^2}{2}} e^{-(x^2+1)t^2} dt$$

Since I'm trying to compute its second order derivative w.r.t. $x$, after applying the rule for differentiation in the integral sign once, when I try to apply it for the second time, the argument of the integral goes to infinity... Can anyone help? Maybe there's some trivial substitution that I'm missing... Even if this differentiation doesn't seems trivial at all, at least for me...

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3 Answers 3

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I have no idea how you did this that would make your integrand go infinite.

$$f\left(x\right)\equiv\int_{-\infty}^{\frac{x^{2}}{2}}e^{-\left(x^{2}+1\right)t^{2}}dt$$

$$\frac{df}{dx}=xe^{-\left(x^{2}+1\right)\left(x^{2}/2\right)^{2}}+\int_{-\infty}^{\frac{x^{2}}{2}}-2xt^{2}e^{-\left(x^{2}+1\right)t^{2}}dt$$ $$=xe^{-\left(x^{6}+x^{4}\right)/4}+\int_{-\infty}^{\frac{x^{2}}{2}}-2xt^{2}e^{-\left(x^{2}+1\right)t^{2}}dt$$

$$\frac{d^{2}f}{dx^{2}}=\left(1-x\frac{6x^{5}+4x^{3}}{4}\right)e^{-\left(x^{2}+1\right)x^{4}/4}+x\left(-2x\left(\frac{x^{2}}{2}\right)^{2}e^{-\left(x^{2}+1\right)x^{4}/4}\right)+\int_{-\infty}^{\frac{x^{2}}{2}}\left(-2t^{2}e^{-\left(x^{2}+1\right)t^{2}}+4x^{2}t^{4}e^{-\left(x^{2}+1\right)t^{2}}\right)dt$$

$$=\left(1-x^{4}-2x^{6}\right)e^{-\left(x^{2}+1\right)x^{4}/4}+\int_{-\infty}^{\frac{x^{2}}{2}}\left(-2t^{2}+4x^{2}t^{4}\right)e^{-\left(x^{2}+1\right)t^{2}}dt$$

This integrand does not go infinite. Are you worried about the $t^4$ as $x\rightarrow-\infty$? Don't be. The exponential in guaranteed to shrink at large t faster than $t^n$ grows for any finite n. So your integrand actually goes to 0 at large t.

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  • $\begingroup$ How can you bring $-2t^2$ outside the integral sign? $\endgroup$
    – james42
    Sep 14, 2015 at 16:12
  • $\begingroup$ @ale42 My bad. I fixed it. Thanks. Good catch. $\endgroup$ Sep 14, 2015 at 22:53
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Let the integral be $G(x)$.

A cheap trick would be to set $u=t\sqrt{1+x^2}$: then the integrand is independent of $x$, and the limits become $-\infty$ and $\frac{x^2}{2}\sqrt{1+x^2}$. You then have $$ G(x) = \frac{1}{\sqrt{1+x^2}} \int_{-\infty}^{x^2\sqrt{1+x^2}/2} e^{-u^2} \, du = \frac{1}{\sqrt{1+x^2}} F(x^2\sqrt{1+x^2}/2) $$ as the integral, where $F(a) = \int_{-\infty}^a e^{-u^2} \, du$.


If you'd rather do it the normal way, you should find for the first derivative $$ e^{-(x^2+1)x^2/2} - \int_0^{x^2/2} 2x t^2 e^{-(x^2+1)t^2} \,dt. $$ Can you manage now?

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By Leibniz integral rule:

$$\frac{d}{dx}\int_{-\infty}^{\frac{x^2}{2}} e^{-(x^2+1)t^2} dt$$

$$= \left[\frac{d}{dx}\int_{-\infty}^{\frac{x^2}{2}} e^{-(a^2+1)t^2} dt\right]|_{a=x} + \int_{-\infty}^{\frac{x^2}{2}} \frac{\partial}{\partial x}e^{-(x^2+1)t^2} dt$$

where

$$\frac{d}{dx}\int_{-\infty}^{\frac{x^2}{2}} e^{-(a^2+1)t^2} dt = e^{-(a^2+1)(\frac{x^2}{2})^2} \frac{d}{dx}\frac{x^2}{2}$$

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