Integration by Parts

Let

$$I= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{x^2\cos x}{1+\pi^x} + \frac{1+\sin^2x}{1+e^{\sin x^{2023}}} \right)\,dx$$

Write

$$I=I_1+I_2$$

where

$$I_1= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2\cos x}{1+\pi^x}\,dx$$ and $$I_2= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^2x}{1+e^{\sin x^{2023}}}\,dx$$

Using

$$\int_{-a}^{a}f(x)\,dx = \int_{0}^{a}[f(x)+f(-x)]\,dx,$$

we get

$$I_1=\int_{0}^{\frac{\pi}{2}}\left(\frac{x^2\cos x}{1+\pi^x}+\frac{x^2\cos x{1+\pi^{-x}}\right)\,dx$$

Since

$$\frac1{1+\pi^x}+\frac1{1+\pi^{-x}}=1,$$

therefore

$$I_1=\int_{0}^{\frac{\pi}{2}}x^2\cos x\,dx$$

Integrating by parts,

$$I_1=\left[x^2\sin x+2x\cos x-2\sin x\right]_0^{\frac{\pi}{2}}$$

$$=\frac{\pi^2}{4}-2$$

Now,

$$I_2= \int_{0}^{\frac{\pi}{2}} (1+\sin^2x)\,dx$$

because

$$\frac1{1+e^{\sin x^{2023}}} + \frac1{1+e^{-\sin x^{2023}}}=1$$

Hence,

$$I_2= \frac{\pi}{2}+\frac{\pi}{4}$$

$$=\frac{3\pi}{4}$$

Therefore,

$$I= \frac{\pi^2}{4}-2+\frac{3\pi}{4}$$

$$= \frac{\pi}{4}(\pi+3)-2$$

Comparing with

$$\frac{\pi}{4}(\pi+a)-2,$$

we get

$$a=3$$

Step 1: Determine the explicit form of f(x)

We are given the functional equation:

$$f(x + 1) = e \cdot f(x)$$

Since f(0) = 1, we can observe the progression for integer values:

$$f(1) = e \cdot f(0) = e^1$$

$$f(2) = e \cdot f(1) = e^2$$

For any x belonging to R, the continuous differentiable function that satisfies this scaling relation along with the initial condition is:

$$f(x) = e^x$$

Substituting this into our target integral gives:

$$I = \int_{0}^{2} [x] \cdot x \cdot e^x \, dx$$

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Step 2: Interval Splitting for the Piecewise Component [x]

The greatest integer function [x] changes value at integer steps. We split the integration limits from 0 to 2 into two sub-intervals: [0, 1) and [1, 2].

$$I = \int_{0}^{1} [x] \cdot x \cdot e^x \, dx + \int_{1}^{2} [x] \cdot x \cdot e^x \, dx$$

Over these intervals, the value of [x] is:

For x in [0, 1), [x] = 0

For x in [1, 2), [x] = 1

Substituting these values into the integrals:

$$I = \int_{0}^{1} (0) \cdot x \cdot e^x \, dx + \int_{1}^{2} (1) \cdot x \cdot e^x \, dx$$

$$I = 0 + \int_{1}^{2} x e^x \, dx$$

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Step 3: Apply Integration by Parts

To evaluate the remaining definite integral, we use integration by parts:

$$\int u \, dv = uv - \int v \, du$$

Let:

$$u = x \implies du = dx$$

$$dv = e^x \, dx \implies v = e^x$$

Applying the boundaries:

$$\int_{1}^{2} x e^x \, dx = \left[ x e^x \right]_{1}^{2} - \int_{1}^{2} e^x \, dx$$

Evaluating the first part at the upper and lower limits:

$$\left[ x e^x \right]_{1}^{2} = (2 \cdot e^2) - (1 \cdot e^1) = 2e^2 - e$$

Evaluating the second part:

$$\int_{1}^{2} e^x \, dx = \left[ e^x \right]_{1}^{2} = e^2 - e$$

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Step 4: Combine the Evaluated Terms

Subtracting the second term from the first term gives:

$$I = (2e^2 - e) - (e^2 - e)$$

$$I = 2e^2 - e - e^2 + e$$

$$I = e^2$$

$$I = \int_{1/2}^{2} \frac{f(x)}{x} \, dx$$

Using the definite integral property under the transformation $$x = \frac{1}{t} \implies dx = -\frac{1}{t^2} dt$$:

When $$x = \frac{1}{2} \implies t = 2$$

When $$x = 2 \implies t = \frac{1}{2}$$

Substituting these into the integral expression:

$$I = \int_{2}^{1/2} \frac{f(1/t)}{1/t} \left(-\frac{1}{t^2}\right) dt = \int_{1/2}^{2} \frac{f(1/t)}{t} \, dt = \int_{1/2}^{2} \frac{f(1/x)}{x} \, dx$$

Adding the two equivalent expressions for $$I$$ gives:

$$2I = \int_{1/2}^{2} \frac{f(x) + f(1/x)}{x} \, dx$$

Substitute the given functional condition into the integral:

$$2I = \int_{1/2}^{2} \frac{\ln(1+x^2)}{x^2} \, dx$$

Now, evaluate the integral using Integration by Parts:

Let $$u = \ln(1+x^2) \implies du = \frac{2x}{1+x^2} dx$$

Let $$dv = \frac{1}{x^2} dx \implies v = -\frac{1}{x}$$

$$\int_{1/2}^{2} \frac{\ln(1+x^2)}{x^2} \, dx = \left[ -\frac{\ln(1+x^2)}{x} \right]_{1/2}^{2} - \int_{1/2}^{2} \left(-\frac{1}{x}\right) \frac{2x}{1+x^2} \, dx$$

$$= \left( -\frac{\ln 5}{2} + 2\ln\left(\frac{5}{4}\right) \right) + 2 \int_{1/2}^{2} \frac{1}{1+x^2} \, dx$$

$$= -\frac{\ln 5}{2} + 2\ln 5 - 4\ln 2 + 2 \left[ \tan^{-1}x \right]_{1/2}^{2}$$

$$= \frac{3}{2}\ln 5 - 4\ln 2 + 2 \left( \tan^{-1}2 - \tan^{-1}\left(\frac{1}{2}\right) \right)$$

Using the identity $$\tan^{-1}2 - \tan^{-1}(1/2) = \tan^{-1}\left(\frac{2 - 1/2}{1 + 2(1/2)}\right) = \tan^{-1}\left(\frac{3}{4}\right) =  \cot^{-1}\left(\frac{4}{3}\right)$$.

$$2I = \frac{3}{2}\ln 5 - 4\ln 2 + 2 \cot^{-1}\left(\frac{4}{3}\right)$$

$$I = \frac{3}{4}\ln 5 - 2\ln 2 +  \cot^{-1}\left(\frac{4}{3}\right)$$

By comparing ,we get $$a = 3 , b = 4, c = 2$

$$ a^2 + b^2 + c^2 = 3^2 + 4^2 + 2^2 = 29$$