## Quantophile

### Financial programming for Quants

#### Category: Calculus

Let $$u$$ and $$v$$ be differentiable functions of $$x$$. Then the differential of the product $$uv$$ is found by the product rule

$$d(uv)=udv+vdu$$

Whence, by integration, we have

$$uv=\int{udv}+\int{vdu}$$

$$\bbox[5pt, border:2px solid blue] { \int{udv}=uv-\int{vdu}\qquad{ (1) } }$$

This formula is called integration by parts. It is often used to integrate expressions, that may be represented as a product of two factors $$u$$ and $$dv$$, in such a way that the finding of the function $$v$$ from its differential $$dv$$ and the evaluation of the integral $$\int{vdu}$$ taken together, be a simpler problem than the direct evaluation of the integral $$\int{udv}$$. To become skilled at breaking up a given element of integration into factors $$u$$ and $$dv$$, one has to solve problems.

Let us consider some functions of a quadratic trinomial. We express the trinomial as a sum or difference of squares and then proceed to integrate the given function.

Example. \begin{aligned} \int{\frac{dx}{x^{2}+2x+5}} \end{aligned}

Solution. We can simplify the quadratic trinomial $$x^{2}+2x+5$$ as,

\begin{aligned} x^{2}+2x+5&=x^{2}+2(x)(1)+1^{2}+4\\ &=(x+1)^{2}+2^{2} \end{aligned}

We make the substitution $$x+1=t$$.

\begin{aligned} I&=\int{\frac{dx}{(x+1)^{2}+2^{2}}}\\ &=\int{\frac{dt}{t^{2}+2^{2}}}\\ &=\frac{1}{2}\arctan\left(\frac{t}{2}\right)+C\\ &=\frac{1}{2}\arctan\left(\frac{x+1}{2}\right)+C \end{aligned}

Some more examples follow.

Let us attempt to graph functions – draw a rough sketch of the function $$y=f(x)$$ or the implicit functions $$F(x,y)=0$$ by hand. These are my notes.

The method of substitution is one of the basic methods for calculating indefinite integrals. Even when we integrate by some other technique, we often resort to substitution in the intermediate stages. The success of integration largely depends on how appropriate the substitution is, in simplifying the integrand. Essentially, the study of the methods of integration, reduces to finding out what kind of substitution should be done. I solve a few interesting problems here.

##### Change of variable

Theorem. The form of the integration formula is independent of the nature of the variable of integration.

Proof. Let it be required to find the integral

$$\int{f(x)dx}$$

Let us change the variable in the expression under the integral sign, putting

$$x=\phi(t)$$

$$\int{f(x)dx}=\int{f[\phi(t)]d(\phi(t))}=\int{f[\phi(t)]\phi'(t)dt}$$

In many problems, a substitution such as the one above, leads to a simpler integral. Let us establish, that the indefinite integral is independent of the nature of the variable of integration – whether $$x$$ or $$\phi(t)$$. It is necessary to prove that their derivatives with respect to $$x$$ are equal.

Differentiating the left side with respect to $$x$$:

$$\left(\int{f(x)dx}\right)_{x}=f(x)$$

Differentiating the right side with respect to $$x$$:

$$\left(\int{f[\phi(t)]\phi'(t)dt}\right)_{x}=\left(\int{f[\phi(t)]\phi'(t)dt}\right)_{t}\cdot\frac{dt}{dx}=f[\phi(t)]\phi'(t)\frac{dt}{dx}$$

But,

$$\frac{dt}{dx}=\frac{1}{dx/dt}=\frac{1}{\phi'(t)}$$

Therefore,

$$\left(\int{f[\phi(t)]\phi'(t)dt}\right)_{x}=f[\phi(t)]\phi'(t)\frac{1}{\phi'(t)}=f[\phi(t)]=f(x)$$

Therefore, the derivatives, with repsect to $$x$$ of the right and left sides are equal as required. Hence, the expressions to the right and left side are the same.