# Is it possible to calculate this integral?

I am new to integrals and I am trying to compute this one:

$$\int \frac{e^-{\frac{\left[ln {T} - \left(\beta-\tau\right)\right]^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}} \,d\tau$$

Note that the integrand function is the PDF of the lognormal distribution multiplied by $$T$$, thus losing the $$T$$ in the denominator because it cancels out with the multiplying $$T$$. Furthermore, the mean has been split into two terms, $$\beta$$ and $$\tau$$.

Assuming that the values of $$\beta$$ and $$\sigma$$ are fixed to be, say, $$11.26$$ and $$1.2$$ respectively, and assuming that both $$T$$ and $$\tau$$ range from $$-\infty$$ to $$\infty$$, is this function integrable?

• I don't think there is an antiderivative for that integrand because $e^{-x^2}$ doesn't have an antiderivative. Sep 24, 2022 at 8:28
• Do you really need the indefinite integral? Sep 24, 2022 at 8:53

Simplifying the integrand gives $$\require{\cancel}$$
$$\int \frac{e^-{\frac{\left[\ln(T) - \left(\beta-\tau\right)\right]^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}} \,\mathrm{d}\tau \overset{\color{darkblue}{u =\frac{\ln(T) - \left(\beta-\tau\right)}{\sigma\sqrt{2}} }}{=} \frac{\cancel{\color{darkblue}{\sigma \sqrt{2}}}}{\cancel{\sigma}\sqrt{\cancel{2}\pi}} \int e^{-u^2}\,\mathrm{d}u$$ Without limits of integration, the problem reduces to finding an antiderivative for $$e^{-x^2}$$. This is impossible using elementary functions because the function is transcendental. Because of this, we defined the error function to be a special function such that $$\frac{1}{\sqrt{\pi}} \int e^{-x^2} \, \mathrm{d}x = \frac{1}{2}\mathrm{erf} \left(x\right) + C$$ so we can conclude $$\int \frac{e^-{\frac{\left[\ln(T) - \left(\beta-\tau\right)\right]^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}} \,\mathrm{d}\tau =\frac{1}{2}\mathrm{erf} \left(\frac{\ln(T) - \left(\beta-\tau\right)}{\sigma\sqrt{2}}\right) + C$$ Alternatively, since you say $$\tau \in (-\infty, \infty)$$ you may also want the definite integral $$\int_{-\infty}^{\infty} \frac{e^-{\frac{\left[\ln(T) - \left(\beta-\tau\right)\right]^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}} \,\mathrm{d}\tau \overset{u =\frac{\ln(T) - \left(\beta-\tau\right)}{\sigma\sqrt{2}} }{=} \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} e^{-u^2}\,\mathrm{d}u =1$$ where on the last step we used that $$\int_{-\infty}^{\infty}e^{-x^2}\mathrm{d}x = \sqrt{\pi}$$, verifying the expected behaviour of a PDF.