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Synopsis:

I cannot duplicate the answer given in a very reputable online symbolic integral calculator as shown in this link ($x$ is $\tau$) although my answer does appear very similar. This tells me that my method is correct but I am making another kind of error -- perhaps in my integration?

Given:

Find the convolution $h(t)$ of $f(t)=t^2$ and $g(t)=e^{at}$. For reference the definition of the convolution and integration by parts are given here...

$$h(t)=(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$$

$$\int udv=uv-\int vdu$$

My Solution:

Begin by inserting the functions into the formula and simplifying...

$$h(t)=t^2*e^{at}=\int_0^t\tau^2\cdot e^{a(t-\tau)}d\tau=e^{at}\int_0^t\tau^2\cdot e^{-\tau}d\tau$$

Now the tedious work begins as we shall need to apply integration by parts twice to reduce $\tau^2$ in the integrand to $1$...

$$u_1=\tau^2\implies du_1=2\tau d\tau$$

$$dv_1=e^{-a\tau}d\tau\implies v_1=\frac{-e^{-a\tau}}{a}$$

$$h(t)=e^{at}\left[-\tau^2\cdot\frac{e^{-a \tau}}{a}\right]_0^t-2e^{at}\int_0^t\tau\cdot \frac{e^{-a\tau}}{-a}d\tau$$

$$=-\frac{t^2}{a}+\frac{2e^{at}}{a}\int_0^t\tau\cdot e^{-a\tau}d\tau$$

We just need to do one more integration by parts...

$$u_2=\tau\implies du_2=d\tau$$

$$dv_2=e^{-a\tau}d\tau\implies v_2=\frac{-e^{-a\tau}}{a}$$

$$h(t)=-\frac{t^2}{a}+\frac{2e^{at}}{a}\left[-\tau\cdot\frac{e^{-a \tau}}{a}\right]_0^t-\frac{2e^{at}}{a}\int_0^t\frac{e^{-a\tau}}{-a}d\tau$$

$$=-\frac{t^2}{a}-\frac{2t}{a^2}+\frac{2e^{at}}{a^2}\int_0^t e^{-a\tau}d\tau$$

$$=-\frac{t^2}{a}-\frac{2t}{a^2}+\frac{2e^{at}}{a^2}\left[\frac{-e^{-a \tau}}{a}\right]_0^t$$

$$=-\frac{t^2}{a}-\frac{2t}{a^2}-\frac{2}{a^3}+\frac{2e^{at}}{a^3}$$

Answer in Link:

$$=\frac{2e^{at}}{a^3}-\frac{t^2}{a}-\frac{2t}{a^2}-\frac{2}{a^3}$$

Question:

See bottom.

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1 Answer 1

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I must be on a good streak. This is now the third post I put up in which the correct solution came to me as I typed it up.

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