Hello all! I am currently trying to express a function (as given below) in terms of the error function (also given below) using substitution. I have no problem using substitution to find integrals but I'm completely stuck on this one. Any and all help would be appreciated. Two "hints" were given: 1) The error function can be composed with other functions. For example, erf x^2 looks like erf x except that the upper limit of the integral is x^2 instead of x. 2) The key to finding the right substitution in this problem is to focus on getting rid of the t^(1/2) in the denominator while ignoring what happens to the exponential factor
Calculus: Substitution
- Thread starter ghof
- Start date
\(\displaystyle \displaystyle f(x) = \int_0^x \dfrac{e^t}{\sqrt{t}} dt \)
You can get rid of the square root in the denominator if it becomes part of du. That is, if du ~ t/sqrt(t). Have you tried u = sqrt(t) ?
So: \(\displaystyle u=\sqrt{t}\), and \(\displaystyle du=\dfrac{1}{2\sqrt{t}}dt \iff dt=2\sqrt{t}du = 2udu\)
\(\displaystyle \displaystyle \int_{0}^x \dfrac{e^{-t}}{\sqrt{t}} dt = \int_{0}^{\sqrt{x}} \dfrac{e^{-u^2}}{u} 2udu\)
\(\displaystyle \displaystyle = 2\int_{0}^{\sqrt{x}} e^{-u^2}du = \sqrt{\pi}\cdot \dfrac{2}{\sqrt{\pi}}\int_{0}^{\sqrt{x}} e^{-u^2}du = \sqrt{\pi}\cdot \text{erf}(\sqrt{x})\)
\(\displaystyle \displaystyle \int_{0}^x \dfrac{e^{-t}}{\sqrt{t}} dt = \int_{0}^{\sqrt{x}} \dfrac{e^{-u^2}}{u} 2udu\)
\(\displaystyle \displaystyle = 2\int_{0}^{\sqrt{x}} e^{-u^2}du = \sqrt{\pi}\cdot \dfrac{2}{\sqrt{\pi}}\int_{0}^{\sqrt{x}} e^{-u^2}du = \sqrt{\pi}\cdot \text{erf}(\sqrt{x})\)