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So this sounds ridiculous I know because $\infty$ is a concept not a number. But hear me out on this one.

Take the $\operatorname*{Si}(x)$ function which is $\int_{0}^{x}\frac{\sin t}{t}~\text{d}t$ obviously this can't be integrated however we know $\lim_{a\to\infty}\operatorname*{Si}(a)=\frac{\pi}{2}$ so we have a point $(a,\frac{\pi}{2})$ to find the slope we just utilize the second fundamental theorem $\operatorname*{Si}'(x)=\frac{\sin x}{x}$ and so the slope is $\lim_{a\to\infty}\operatorname*{Si}'(a)=0$ so we have a point and a slope so we end up with a tangent line at $\infty$ being $y=\frac{\pi}{2}$.

Is all of this even valid? I don't think I've broken calculus rules and if I haven't could I do this for other functions that can't be integrated such as $e^{-x^2}$ or $\frac{\tan^{-1}(ax)-\tan^{-1}(bx)}{x}$(for constants $a$ and $b$)? Their integrals can both be evaluated at $\infty$. I know it's a long shot but thanks in advance.

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    There is a terminology issue: *integrable* does not imply that the primitive is a "nice" function. $e^{-x^2}$ is integrable, $\frac{\sin x}{x}$ is not Lebesgue-integrable over $\mathbb{R}^+$ but it is improperly Riemann-integrable. – 2017-01-23
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    You've noticed that the function $\operatorname{Si}(a)$ has a horizontal [asymptote](https://en.wikipedia.org/wiki/Asymptote) $y = \pi/2$. Many functions have horizontal asymptotes. In particular, if $\int_0^\infty f(t)\,dt$ exists then the function $F(x) = \int_0^x f(t)\,dt$ will have a horizontal asymptote $y = \int_0^\infty f(t)\,dt$. – 2017-01-23
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    @AntonioVargas yeah thank you,I actually just realized that and then saw this comment. – 2017-01-23
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    $\infty$ from calculus is a number. The relevant number system is the extended reals. (note that this number system isn't as nice for doing algebra as the ordinary reals, but it's an excellent number system for doing arithmetic with limits) – 2017-01-25

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Actually, you have, by definition, a horizontal asymptote. Of course, if the limit of the derivative was non-zero, then you'd have a linear asymptote:

$$f(x)\sim ax$$

Where $f\sim g$ means $\lim\limits_{x\to\infty}\frac fg=1$

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    Doesn't it change it matter that it's in an integral, for example I could approximate the values of the integral using the line – 2017-01-23
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    @TehRod oh, you want to approximate an integral? I'd look up Taylor's theorem if you haven't. – 2017-01-23