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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?

J. Doe
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1 Answers1

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The function is indeed integrable, and you can give a closed form for its antiderivative and its definite integral.

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.

Robert Lee
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