I did this calculation and something just isn't quite right. I am getting the result $$\int_{0}^{\infty}e^{-x^2}\:dx=\frac{-3\sqrt{\pi}}{4}$$ which is absolutely wrong. I would be grateful if anyone could help me find the mistake in my working.
I assumed $$I(a)=\int_{0}^{\infty}e^{-ax^2}\:dx\;\;\;\;\;\; (a>0)$$ $$\Rightarrow \frac{d}{da}(I(a))=\int_{0}^{\infty}(-x^2)e^{-ax^2}\:dx$$ Then I substituted $x^2=t\Rightarrow dx=\frac{1}{2\sqrt{t}}dt$.
$$\Rightarrow \frac{d}{da}(I(a))=\left(\frac{-1}{2}\right)\int_{0}^{\infty}e^{-at}\sqrt{t}\:dt\;\;\;\;\;\; -(i)$$ $$\Rightarrow \frac{d^2}{da^2}(I(a))=\left(\frac{1}{2}\right)\int_{0}^{\infty}e^{-at}(t\sqrt{t})\:dt\;\;\;\;\;\; -(ii)$$
I applied Integration by Parts on $(i)$ took $u=e^{-at}$ and $v=\frac{2t\sqrt{t}}{3}$.
$$\int u\:dv=uv\;-\int v\:du\Rightarrow \int_{0}^{\infty}e^{-at}\sqrt{t}\:dt=\left[e^{-at}\cdot\frac{2t\sqrt{t}}{3}\right]_{0}^{\infty}+\left(\frac{2a}{3}\right)\int_{0}^{\infty}e^{-at}\:t\sqrt{t}\:dt\;\;\;\;\;\; -(iii)$$ Using $(i),(ii)$ and $(iii)$. $$\Rightarrow \left(\frac{-1}{2}\right)\int_{0}^{\infty}e^{-at}\sqrt{t}\:dt=\left(\frac{-a}{3}\right)\int_{0}^{\infty}e^{-at}\:t\sqrt{t}\:dt=\left(\frac{-2a}{3}\right)\frac{d^2}{da^2}(I(a))=\frac{d}{da}(I(a))$$ $$\Rightarrow \frac{d}{da}(I(a))=Ke^{\frac{-2a}{3}}+C$$
We know $\lim\limits_{a\to \infty}(\frac{d}{da}(I(a)))=0$ and $\frac{d}{da}(I(a))|_{a=1}=\frac{\sqrt{\pi}}{2}.$ $$\Rightarrow \frac{d}{da}(I(a))=\left(\frac{e^{\frac{2}{3}}\sqrt{\pi}}{2}\right)e^{\frac{-2a}{3}}\Rightarrow I(a)=\left(\frac{(-3)\:e^{\frac{2}{3}}\sqrt{\pi}}{4}\right)e^{\frac{-2a}{3}}+C_1$$ We also know that $\lim\limits_{a\to\infty}(I(a))=0$. $$\Rightarrow I(a)=\left(\frac{(-3)\:e^{\frac{2}{3}}\sqrt{\pi}}{4}\right)e^{\frac{-2a}{3}}$$ $$\Rightarrow I(1)=\int_{0}^{\infty}e^{-x^2}\:dx=\frac{-3\sqrt{\pi}}{4}$$
But this result is absolutely wrong! Please help me correct my mistake. Thank You!
From$$\left(\frac{-2a}{3}\right)\frac{\mathrm d^2}{\mathrm da^2}(I(a))=\frac{\mathrm d}{\mathrm da}(I(a)),$$you get that$$\frac{\mathrm d^2}{\mathrm da^2}(I(a))=-\frac3{2a}\frac{\mathrm d}{\mathrm da}(I(a)),$$and from this you cannot deduce that$$\frac{\mathrm d}{\mathrm da}(I(a))=Ke^{-2a/3}+C.\tag1$$In fact, if you do have $(1)$ for some constants $K$ and $C$, then\begin{align}\frac{\mathrm d^2}{\mathrm da^2}(I(a))&=\frac{\mathrm d}{\mathrm da}\left(Ke^{-2a/3}+C\right)\\&=-\frac23Ke^{-2a/3}\\&\ne-\frac3{2a}\left(Ke^{-2a/3}+C\right)\\&=-\frac3{2a}\frac{\mathrm d}{\mathrm da}(I(a)).\end{align}